**Q-1: Check whether the following are the probability distributions of a random variable or not. Also, justify your answer.**

**(i) **

X |
0 |
1 |
2 |

P(x) |
0.3 |
0.5 |
0.2 |

** **

**(ii)**

X |
0 |
1 |
2 |
3 |
4 |

P(X) |
0.2 |
0.6 |
0.3 |
-0.2 |
0.3 |

** **

**(iii) **

Y |
0 |
1 |
2 |

P(Y) |
0.7 |
0.2 |
0.3 |

** **

**(iv) **

Z |
3 |
2 |
1 |
0 |
-1 |

P(Z) |
0.4 |
0.3 |
0.5 |
0.2 |
0.06 |

** **

**Solution:**

We know that the sum of all of the probabilities in the probability distribution should be one.

(i)

X | 0 | 1 | 2 |

P(x) | 0.3 | 0.5 | 0.2 |

Sum of the probabilities = 0.3 + 0.5 + 0.2 = 1.0

Hence, this given table is a probability distribution of the random variables.

(ii)

X | 0 | 1 | 2 | 3 | 4 |

P(X) | 0.2 | 0.6 | 0.3 | -0.2 | 0.3 |

Here, from the table we can see that for X = 3, P(X) = -0.2

We know that, the probability for any of the observation cannot be negative.

Hence, the table given is not a probability distribution for the random variables.

(iii)

Y | 0 | 1 | 2 |

P(Y) | 0.7 | 0.2 | 0.3 |

Sum of all the probabilities given in the Q. = 0.7 + 0.2 + 0.3 = 1.2 ≠ 1

**Hence, the table given is not a probability distribution for the random variables.**

**(iv) **

Z | 3 | 2 | 1 | 0 | -1 |

P(Z) | 0.4 | 0.3 | 0.5 | 0.2 | 0.06 |

**Here, sum of all of the probabilities given in the Q. = 0.4 + 0.3 + 0.5 + 0.2 + 0.06 = 1.46 ****≠**** 1**

Hence, the table given is not a probability distribution for the random variables.

** **

**Q.2: There are 5 red balls and 2 black balls. From that, two balls are drawn at random. Let X represents the number of the black balls. Find the values for X. Check whether X is a random variable or not?**

** **

**Solution: **

Let, the two balls drawn randomly are represented as RR, RB, BR, and BB, where R is for a black ball and B is for a red ball.

**According to Q,**

X represents the number of the black balls.

Thus,

**X (RR) = 0**

**X (RB) = 1**

**X (BR) = 1**

**X (BB) = 2**

**Hence, the possible values of X are 0, 1 and 2.**

**Therefore, C is a random variable.**

** **

**Q.3: Let us consider X which represents the difference between the total number of tails and the number of heads which can be obtained when a single coin is tossed for 5 times. Find the possible values for X.**

**Solution:**

** As per the data given in the Q, **

A coin will be tossed for 5 times and every time the result is observed. Also, X represents the difference between the total number of heads and tails observed after each toss.

Thus,

X (5H, 0T) = | 5 – 0 | **= 5**

X (4H, 1T) = | 4 – 1 | **= 3**

X (3H, 2 T) = | 3 – 2 | **= 1**

X(2H, 3T) = | 2 – 3 | **= 1**

X(1H, 4T) = | 1 – 4 | **= 3**

X(0H, 5T) = | 0 – 5 | **= 5**

**Hence, the possible values for X are 5, 3 and 1.**

** **

**Q.4: What will be the probability distribution for?**

**(i) Number of tails after tossing a coin for twice.**

**(ii) Number of heads after simultaneously tossing a coin for three times.**

**(iii) Number of tails after tossing a coin for four times.**

** **

**Solution:**

**(i) Once a coin is tossed for two times, the chances are:**

{TT, TH, HT, HH}, where H is for heads and T is for tails.

Let, the total number of tails be represented by X.

Then,

**X(TT) = 2, X(TH) = 1, X(HT) = 1 and X(HH) = 0.**

Hence, X can take the values of 0, 1 and 2.

We know that,

**P(TT) =** P(TH) = P(HT) = P(HH) = \(\frac{ 1 }{ 4 }\)

P(X = 0) = P(HH) = \(\frac{ 1 }{ 4 }\)

**P(X = 1) =** P (HT) + P(TH) = \(\frac{ 1 }{ 4 } + \frac{ 1 }{ 4 }\) = \(\frac{ 1 + 1 }{ 4 } = \frac{ 2 }{ 4 } = \frac{ 1 }{ 2 }\)

**P(X = 2) =** P(TT) = \(\frac{ 1 }{ 4 }\)

**Hence, the probability distribution which is required as per the Q.’s demand is given below:**

X | 0 | 1 | 2 |

P(X) | \(\frac{ 1 }{ 4 }\) | \(\frac{ 1 }{ 2 }\) | \(\frac{ 1 }{ 4 }\) |

** **

**(ii) Once a coin is tossed for three times, the chances are:**

{TTT, TTH, THH, HHH, HHT, HTT, HTH, THT}, where H is for heads and T is for tails.

Let, the total number of tails be represented by X.

Then,

X(HHH) = 3, X(HTH) = 2, X(HHT) = 2, X(THH) = 2, X(TTH) = 1, X(HTT) = 1, X(THT) = 1 and X(TTT) = 0.

**Hence, X can take the values of 0, 1, 2 and 3.**

Now,

**P(X = 0) = **P(TTT) = \(\frac{ 1 }{ 8 }\)

**P(X = 1) = **P (TTH) + P(HTT) + P(THT) = \(\frac{ 1 }{ 8 } + \frac{ 1 }{ 8 } + \frac{ 1 }{ 8 }\) = \(\frac{ 1 + 1 + 1 }{ 8 } = \frac{ 3 }{ 8 } \)

**P(X = 2) = **P (HTH) + P(THH) + P(HHT) = \(\frac{ 1 }{ 8 } + \frac{ 1 }{ 8 } + \frac{ 1 }{ 8 }\) = \(\frac{ 1 + 1 + 1 }{ 8 } = \frac{ 3 }{ 8 } \)

**P(X = 3) = **P(HHH) = \(\frac{ 1 }{ 8 }\)

**Hence, the probability distribution which is required as per the Q.’s demand is given below:**

X | 0 | 1 | 2 | 3 |

P(X) | \(\frac{ 1 }{ 8 }\) | \(\frac{ 3 }{ 8 }\) | \(\frac{ 3 }{ 8 }\) | \(\frac{ 1 }{ 8 }\) |

**(iii) Once a coin is tossed for four times, the chances are:**

{TTTT, TTTH, TTHT, TTHH, THTT, THTH, THHT, THHH, HTTT, HTTH, HTHH, HTHT, HHTT, HHTH, HHHT, HHHH}, where H is for heads and T is for tails.

Let, the total number of tails be represented by X.

Then,

X(TTTT) = 4, X(TTTH) = 3, X(TTHT) = 3, X(THTT) = 3, X(HTTT) = 3, X(TTHH) = 2, X(THTH) = 2, X(THHT) = 2, X(HTTH) = 2, X(HTHT) = 2, X(HHTT) = 2, X(THHH) = 1, X(HTHH) = 1, X(HHTH) = 1, X(HHHT) = 1 and X(HHHH) = 0.

**Hence, X can take the values of 0, 1, 2, 3 and 4.**

Now,

**P(X = 0) =** P(HHHH) = \(\frac{ 1 }{ 16 }\)

**P(X = 1) =** P (THHH) + P(HTHH) + P(HHTH) + P(HHHT) = \(\frac{ 1 }{ 16 } + \frac{ 1 }{ 16 } + \frac{ 1 }{ 16 } + \frac{ 1 }{ 16 }\) = \(\frac{ 1 + 1 + 1 + 1 }{ 16 } = \frac{ 4 }{ 16 } = \frac{ 1 }{ 4 } \)

**P(X = 2) =** P (TTHH) + P(THTH) + P(THHT) + P(HTTH) + P(HTHT) + P(HHTT) = \(\frac{ 1 }{ 16 } + \frac{ 1 }{ 16 } + \frac{ 1 }{ 16 } + \frac{ 1 }{ 16 } + \frac{ 1 }{ 16 } + \frac{ 1 }{ 16 }\) = \(\frac{ 1 + 1 + 1 + 1 + 1 + 1 }{ 16 } = \frac{ 6 }{ 16 } = \frac{ 3 }{ 8 } \)

**P(X = 3) =** P (TTTH) + P(TTHT) + P(THTT) + P(HTTT) = \(\frac{ 1 }{ 16 } + \frac{ 1 }{ 16 } + \frac{ 1 }{ 16 } + \frac{ 1 }{ 16 }\) = \(\frac{ 1 + 1 + 1 + 1 }{ 16 } = \frac{ 4 }{ 16 } = \frac{ 1 }{ 4 } \)

**P(X = 0) =** P(TTTT ) = \(\frac{ 1 }{ 16 }\)

**Hence, the probability distribution which is required as per the Q.’s demand is given below:**

X | 0 | 1 | 2 | 3 | 4 |

P(X) | \(\frac{ 1 }{ 16 }\) | \(\frac{ 1 }{ 4 }\) | \(\frac{ 3 }{ 8 }\) | \(\frac{ 1 }{ 4 }\) | \(\frac{ 1 }{ 16 }\) |

** **

**Q.5: What will be the probability distribution of the numbers for the two success tosses of a die, where a success will be defined as?**

**(i) Obtained number which is greater than 3.**

**(ii) On the die, six appears for at least once.**

** **

**Solution:**

When we will toss a die for two times, we will obtain at least (6 × 6) = 36 number of the observations.

Let us consider X which is a random variable, which represents total number of successes.

**(i) In this case, numbers which is greater than 3 is referred to as success.**

**P(X = 0) =** P (Number which are less than or equal to 3 after having both the tosses) = \(\frac{ 3 }{ 6 } \times \frac{ 3 }{ 6 } = \frac{ 1 }{ 4 }\)

**P(X = 1) =** P(Numbers which are less than or equal to 3 in first toss and greater than 3 in the second toss) + P(Numbers which are greater than 3 in first toss and less than or equal to 3 in the second toss)

= \(\frac{ 3 }{ 6 } \times \frac{ 3 }{ 6 } + \frac{ 3 }{ 6 } \times \frac{ 3 }{ 6 }\)

= \( \frac{ 1 }{ 2 } \times \frac{ 1 }{ 2 } + \frac{ 1 }{ 2 } \times \frac{ 1 }{ 2 }\)

= \( \frac{ 1 }{ 4 } + \frac{ 1 }{ 4 }\)

= \(\frac{ 1 }{ 2 }\)

P(X = 2) = P(Numbers which are greater than 3 after both toss) = \(\frac{ 3 }{ 6 } \times \frac{ 3 }{ 6 }\)

= \(\frac{ 1 }{ 2 } \times \frac{ 1 }{ 2 }\)

= \(\frac{ 1 }{ 4 }\)

**Hence, the probability distribution which is required as per the Q.’s demand is given below:**

X | 0 | 1 | 2 |

P(X) | \(\frac{ 1 }{ 4 }\) | \(\frac{ 1 }{ 2 }\) | \(\frac{ 1 }{ 4 }\) |

**(ii) In this case, success will be when six appeared at least for once after tossing the dice simultaneously. **

**P(Y = 0) =** P(six doesn’t appeared on either of the dice) = \(\frac{ 5 }{ 6 } \times \frac{ 5 }{ 6 } = \frac{ 25 }{ 36 } \)

P(Y = 1) = P(six appeared at least on one of the dice) = \(\frac{ 1 }{ 6 } \times \frac{ 5 }{ 6 } + \frac{ 5 }{ 6 } \times \frac{ 1 }{ 6 } = \frac{ 5 }{ 36 } + \frac{ 5 }{ 36 } = \frac{ 10 }{ 36 } = \frac{ 5 }{ 18 }\)

**Hence, the probability distribution which is required as per the Q.’s demand is given below:**

Y | 0 | 1 |

P(Y) | \(\frac{ 25 }{ 36 }\) | \(\frac{ 5 }{ 18 }\) |

**Q.6: Consider a lot of 34 bulbs among which 10 bulbs are defective. From that lot, a sample of 3 bulbs is drawn at random with having replacement. What is the probability distribution for the number of the defective bulbs?**

** **

**Solution:**

As per the data given in the Q., we have

A lot of 34 bulbs among which 10 bulbs are defective.

So, the number of non- defective bulbs = 34 – 10** = 24 bulbs**

Now,

A sample of 3 bulbs is drawn at random from the lot with having replacement.

Let us consider X be a random variable which denotes the total number of defective bulbs among the sample drawn.

**P(X = 0) =** P (3 non- defective and 0 defective) = ^{3}C_{0}. \(\frac{ 3 }{ 4 } \times \frac{ 3 }{ 4 } \times \frac{ 3 }{ 4 }\) = \(\frac{ 3! }{ 0! 3! } \times \frac{ 27 }{ 64 }\) = \(1 \times \frac{ 27 }{ 64 } = \frac{ 27 }{ 64 }\)

**P(X = 1) =** P (2 non- defective and 1 defective) = ^{3}C_{1}. \(\frac{ 1 }{ 4 } \times \frac{ 3 }{ 4 } \times \frac{ 3 }{ 4 }\) = \(\frac{ 3! }{ 1! 2! } \times \frac{ 9 }{ 64 }\) = \(\frac{ 3 \times 2! }{ 2! } \times \frac{ 9 }{ 64 } = 3 \times \frac{ 9 } { 64 } = \frac{ 27 }{ 64 }\)

**P(X = 2) =** P (1 non- defective and 2 defective) = ^{3}C_{2}. \(\frac{ 1 }{ 4 } \times \frac{ 1 }{ 4 } \times \frac{ 3 }{ 4 }\) = \(\frac{ 3! }{ 2! 1! } \times \frac{ 3 }{ 64 }\) = \(\frac{ 3 \times 2! }{ 2! } \times \frac{ 3 }{ 64 } = 3 \times \frac{ 3 } { 64 } = \frac{ 9 }{ 64 }\)

P(X = 3) = P (0 non- defective and 3 defective) = ^{3}C_{3}. \(\frac{ 1 }{ 4 } \times \frac{ 1 }{ 4 } \times \frac{ 1 }{ 4 }\) = \(\frac{ 3! }{ 3! 0! } \times \frac{ 1 }{ 64 }\) = \(1 \times \frac{ 1 }{ 64 } = \frac{ 1 }{ 64 }\)

**Hence, the probability distribution which is required as per the Q.’s demand is given below:**

X | 0 | 1 | 2 | 3 |

P(X) | \(\frac{ 27 }{ 64 }\) | \(\frac{ 27 }{ 64 }\) | \(\frac{ 9 }{ 64 }\) | \(\frac{ 1 }{ 64 }\) |

**Q.7: Consider a situation where a coin will be biased in such a manner that the tail will occur likely 4 times the head. What will be the probability distribution of the number of heads, by considering the situation that the coin is tossed for two times?**

** **

**Solution:**

Let us consider that the probability of getting the head in the biased coin is x.

Then,

P(H) = x

⟹ P(T) = 4x

**For the biased coin, P(H) + P(T) = 1**

⟹ x + 4x = 1

⟹ 5x = 1

⟹ x = \(\frac{ 1 }{ 5 }\)

**Hence, P(H) =** \(\frac{ 1 }{ 5 }\) and P(T) = \(\frac{ 4 }{ 5 }\)

When the coin will be tossed for two times, then the sample space will be { TT, TH, HT, HH }

**Let, X be the random variable which represents the number of heads.**

Then,

**P(X = 0) =** P(no heads) = P(T) × P(T) = \(\frac{ 4 }{ 5 } \times \frac{ 4 }{ 5 } = \frac{ 16 }{ 25 }\)

**P(X = 1) =** P(one heads) = P(HT) + P(TH) = \(\frac{ 1 }{ 5 } \times \frac{ 4 }{ 5 } + \frac{ 4 }{ 5 } \times \frac{ 1 }{ 5 } = \frac{ 4 }{ 25 } + \frac{ 4 }{ 25 } = \frac{ 8 }{ 25 }\)

**P(X = 2) =** P(two heads) = P(H) × P(H) = \(\frac{ 1 }{ 5 } \times \frac{ 1 }{ 5 } = \frac{ 1 }{ 25 }\)

**Hence, the probability distribution which is required as per the Q.’s demand is given below:**

X | 0 | 1 | 2 |

P(X) | \(\frac{ 16}{ 25 }\) | \(\frac{ 8 }{ 25 }\) | \(\frac{ 1 }{ 25 }\) |

** **

** **

**Q.8: The following probability distribution is for the random variable X.**

X |
0 |
1 |
2 |
3 |
4 |
5 |
6 |
7 |

P(X) |
0 |
k |
2k |
2k |
3k |
k^{2} |
2k^{2} |
7k^{2} + k |

**Find:**

**(i) The value of k**

**(ii) P(X < 4) **

**(iii) P(X > 5)**

**(iv) P (0 < X < 2)**

** **

**Solution:**

We know that, the sum of all of the probabilities of the probability distribution of a random variable is 1.

Thus,

k + 2k + 2k + 3k + k^{2 }+ 2k^{2} + 7k^{2} + k = 1

⟹ 10k^{2} + 9k = 1

⟹ 10k^{2} + 9k – 1 = 0

⟹ 10k^{2} + 10k – k – 1 = 0

⟹ 10k(k + 1) – 1(k + 1) = 0

⟹ (k + 1)(10k – 1) = 0

⟹ k = -1, \(\frac{ 1 }{ 10 }\)

Now,

k = -1 is not possible because the probability for any of the event cannot ever be negative.

**So, k = \(\frac{ 1 }{ 10 }\)**

**(ii) ****P(X < 4) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) **

= o + k + 2k + 2k = 5k

= \(5 \times \frac{ 1 }{ 10 }\) **= \(\frac{ 1 }{ 2 }\)**

**(iii) P(X > 5) = P(X = 6) + P(X = 7) **

= 2k^{2} + 7k^{2} + k

= 9k^{2} + k

= k (9k + 1)

= \(\frac{ 1 }{ 10 }\)(\( 9 \times \frac{ 1 }{ 10 }\) + 1)

= \(\frac{ 1 }{ 10 }\)(\(\frac{ 9 }{ 10 } + 1 \))

= \(\frac{ 1 }{ 10 }\) (\( \frac{ 9 + 10 }{ 10 }\))

= \(\frac{ 1 }{ 10 }\)(\(\frac{ 19 }{ 10 }\))

= \(\frac{ 19 }{ 100 }\)

**(iv) P(0 < X < 2) = P(X = 1)**

= k = \(\frac{ 1 }{ 10 }\)

**Q.9: Consider P(Y) be the probability distribution of the random variable Y for the following forms, where some number is.**

**P(Y) = **

**2a, if a = 0**

**3a, if a = 1**

**4a, if a = 2**

**0, otherwise**

**(i) Find the value of a.**

**(ii) Determine P(Y < 2), P(Y ≥ 2), and P (Y ≤ 2).**

** **

**Solution:**

**(i) As we know that the sum for all of the probabilities of the probability distribution of random variables is given by 1.**

Thus,

**2a + 3a + 4a + 0 = 1**

⟹ 9a = 1

⟹ a = \(\frac{ 1 }{ 9 }\)

**(ii) P(Y < 2) = P(Y = 0) + P(Y = 1)**

= 2a + 3a

= 5a = \( 5 \times \frac{ 1 }{ 9 }\) = \(\frac{ 5 }{ 9 }\)

**P(Y ≥ 2) = P(Y = 2) + P(Y > 2) **

= 4a + 0

= 4a = \(4 \times \frac{ 1 }{ 9 }\) = \(\frac{ 4 }{ 9 }\)

**P(Y ≤ 2) = P(Y = 2) + P(Y = 1) + P(Y = 0)**

= 4a + 3a + 2a

= 9a = \(9 \times \frac{ 1 }{ 9 }\) **= 1**

**Q.10: After tossing a fair coin, what will be the mean number for tails?**

** **

**Solution:**

Let us consider X which denotes the chances of success to get tails.

Thus,

For this case, the sample space will be

**S = {TTT, TTH, THT, THH, HTT, HTH, HHT, HHH}**

Let, the total number of tails be represented by X.

Then,

X(HHH) = 0, X(HTH) = 1, X(HHT) = 1, X(THH) = 1, X(TTH) = 2, X(HTT) = 2, X(THT) = 2 and X(TTT) = 3.

Here, we can see that X will take values 0, 1, 2 or 3.

Now,

**P(X = 0) = P(HHH) **

= \(\frac{ 1 }{ 2 } \times \frac{ 1 }{ 2 } \times \frac{ 1 }{ 2 }\)

= \(\frac{ 1 }{ 8 }\)

**P(X = 1) = P(HTH) + P(HHT) + P(THH) **

= \(\frac{ 1 }{ 2 } \times \frac{ 1 }{ 2 } \times \frac{ 1 }{ 2 }\) + \(\frac{ 1 }{ 2 } \times \frac{ 1 }{ 2 } \times \frac{ 1 }{ 2 }\) + \(\frac{ 1 }{ 2 } \times \frac{ 1 }{ 2 } \times \frac{ 1 }{ 2 }\)

= \(\frac{ 1 }{ 8 } + \frac{ 1 }{ 8 } + \frac{ 1 }{ 8 }\)

= \(\frac{ 3 }{ 8 }\)

**P(X = 2) = P(TTH) + P(HTT) + P(THT) **

= \(\frac{ 1 }{ 2 } \times \frac{ 1 }{ 2 } \times \frac{ 1 }{ 2 }\) + \(\frac{ 1 }{ 2 } \times \frac{ 1 }{ 2 } \times \frac{ 1 }{ 2 }\) + \(\frac{ 1 }{ 2 } \times \frac{ 1 }{ 2 } \times \frac{ 1 }{ 2 }\)

= \(\frac{ 1 }{ 8 } + \frac{ 1 }{ 8 } + \frac{ 1 }{ 8 }\)

= \(\frac{ 3 }{ 8 }\)

**P(X = 0) = P(TTT) **

= \(\frac{ 1 }{ 2 } \times \frac{ 1 }{ 2 } \times \frac{ 1 }{ 2 }\) = \(\frac{ 1 }{ 8 }\)

**Hence, the probability distribution which is required as per the Q.’s demand is given below:**

X | 0 | 1 | 2 | 3 |

P(X) | \(\frac{ 1 }{ 8 }\) | \(\frac{ 3 }{ 8 }\) | \(\frac{ 3 }{ 8 }\) | \(\frac{ 3 }{ 8 }\) |

**Mean of X E(X), µ = \(\sum X_{ i }P\left (X_{ i } \right)\)**

= \(0 \times \frac{ 1 }{ 8 } + 1 \times \frac{ 3 }{ 8 } + 2 \times \frac{ 3 }{ 8 } + 3 \times \frac{ 1 }{ 8 }\)

= \(\frac{ 3 }{ 8 } + \frac{ 6 }{ 8 } + \frac{ 3 }{ 8 }\)

= \(\frac{ 12 }{ 8 }\)

= \(\frac{ 3 }{ 2 }\) **= 1.5**

** **

**Q.11: Take two dices which are thrown simultaneously. If Y denotes the number of times we will get sixes, then what will be the expectation of Y?**

** **

**Solution:**

As per the data given in the Q., Y represents the total number of sixes which will be obtained after throwing two dice simultaneously.

Thus, Y will take the values of 0, 1 or 2.

Hence,

P(Y = 0) = P (no sixes will be obtained in either of the throw) = \(\frac{ 5 }{ 6 } \times \frac{ 5 }{ 6 } = \frac{ 25 }{ 36 }\)

P(Y = 1) = P (six at the first dice and other number rather than 6 on the second dice) + P(other number rather than six on the first dice and six on the second dice)

= 2 \(\left (\frac{ 5 }{ 6 } + \frac{ 1 }{ 6 } \right)\)

= \( 2 \times \frac{ 5 }{ 36 }\)

= \(\frac{ 5 }{ 18 }\)

P(Y = 2) = P (sixes will be obtained in both of the throw) = \(\frac{ 1 }{ 36 }\)

**Hence, the probability distribution which is required as per the Q.’s demand is given below:**

Y | 0 | 1 | 2 |

P(Y) | \(\frac{ 25}{ 36 }\) | \(\frac{ 10 }{ 36 }\) | \(\frac{ 1 }{ 36 }\) |

**Therefore,**

**Expectation of X = E(X) = \(\sum X_{ i }P\left (X_{ i } \right)\)**

= \(0 \times \frac{ 25 }{ 36 } + 1 \times \frac{ 10 }{ 36 } + 2 \times \frac{ 1 }{ 36 }\)

= \(0 + \frac{ 5 }{ 18 } + \frac{ 1 }{ 18 }\)

= \(\frac{ 6 }{ 18 }\) = \(\frac{ 1 }{ 3 }\)

** **

**Q.12: From the starting six positive integers, two numbers will be selected at random, without any replacement. Let, Y denotes the larger number among those two numbers obtained. Find E(Y).**

** **

**Solution:**

As per the Q.’s demand, from the starting six positive integers, two numbers will be selected, without having replacement which will be done in 6 × 5 = 30 ways.

Let us consider Y which represents the two numbers obtained which are larger. Thus,

**Y will take values 2, 3, 4, 5 or 6. **

**For Y = 2**, the possible observations will be (1, 2) and (2, 1).

Then,

**P(Y = 2) =** \(\frac{ 2 }{ 30 } = \frac{ 1 }{ 15 }\)

**For Y = 3**, the possible observations will be (1, 3), (2, 3), (3, 1) and (3, 2)

Then,

**P(Y = 3) =** \(\frac{ 4 }{ 30 } = \frac{ 2 }{ 15 }\)

For Y = 4, the possible observations will be (1, 4), (2, 4), (3, 4), (4, 1), (4, 2) and (4, 3)

Then,

**P(Y = 4) =** \(\frac{ 6 }{ 30 } = \frac{ 1 }{ 5 }\)

**For Y = 5,** the possible observations will be (1, 5), (2, 5), (3, 5), (4, 5), (5, 4), (5, 3), (5, 2), (5, 1)

Then,

**P(Y = 5) =** \(\frac{ 8 }{ 30 } = \frac{ 4 }{ 15 }\)

For Y = 6, the possible observations will be (1, 6), (2, 6), (3, 6), (4, 6), (5, 6), (6, 5), (6, 4), (6, 3), (6, 2), (6, 1)

Then,

P(Y = 6) = \(\frac{ 10 }{ 30 } = \frac{ 1 }{ 3 }\)

**Hence, the probability distribution which is required as per the Q.’s demand is given below:**

X | 2 | 3 | 4 | 5 | 6 |

P(X) | \(\frac{ 1 }{ 15 }\) | \(\frac{ 2 }{ 15 }\) | \(\frac{ 1 }{ 5 }\) | \(\frac{ 4 }{ 15 }\) | \(\frac{ 1 }{ 3 }\) |

Therefore,

**Expectation of X = E(X) = \(\sum X_{ i }P\left (X_{ i } \right)\)**

= \(2 \times \frac{ 1 }{ 15 } + 3 \times \frac{ 2 }{ 15 } + 4 \times \frac{ 1 }{ 5 } + 5 \times \frac{ 4 }{ 15 } + 6 \times \frac{ 1 }{ 3 }\)

= \(\frac{ 2 }{ 15 } + \frac{ 6 }{ 15 } + \frac{ 4 \times 3 }{ 15 } + \frac{ 20 }{ 15 } + \frac{ 6 \times 5 }{ 15 } \)

= \(\frac{ 2 }{ 15 } + \frac{ 6 }{ 15 } + \frac{ 12 }{ 15 } + \frac{ 20 }{ 15 } + \frac{ 30 }{ 15 } \)

= \(\frac{ 70 }{ 15 }\) = \(\frac{ 14 }{ 3 }\)

** **

**Q.13: Y denotes the sum of all the numbers which are obtained when two fair dice are rolled. What will be the variance and the standard deviation of Y.?**

** **

**Solution:**

Number of observations which are obtained when two dice is rolled is 6 × 6 = 36

**P(Y = 2) =** P(1, 1) = \(\frac{ 1 }{ 36 }\)

**P(Y = 3) =** P(1, 2) + P(2, 1) = \(\frac{ 2 }{ 36 }\)

**P(Y = 4) =** P(1, 3) + P(2, 2) + P(3, 1) = \(\frac{ 3 }{ 36 }\)

**P(Y = 5) =** P(1, 4) + P(2, 3) + P(3, 2) + P(1, 4) = \(\frac{ 4 }{ 36 }\)

**P(Y = 6) =** P(1, 5) + P(2, 4) + P(3, 3) + P(4, 2) + P(5, 1) = \(\frac{ 5 }{ 36 }\)

**P(Y = 7) =** P(1, 6) + P(2, 5) + P(3, 4) + P(4, 3) + P(5, 2) + P(6, 1) = \(\frac{ 6 }{ 36 }\)

**P(Y = 8) =** P(2, 6) + P(3, 5) + P(4, 4) + P(5, 3) + P(6, 2) = \(\frac{ 5 }{ 36 }\)

**P(Y = 9) =** P(3, 6) + P(4, 5) + P(5, 4) + P(3, 6) = \(\frac{ 4 }{ 36 }\)

**P(Y = 10) =** P(4, 6) + P(5, 5) + P(6, 4) = \(\frac{ 3 }{ 36 }\)

**P(Y = 11) =** P(5, 6) + P(6, 5) = \(\frac{ 2 }{ 36 }\)

**P(Y = 12) =** P(6, 6) = \(\frac{ 1 }{ 36 }\)

**Hence, the probability distribution which is required as per the Q.’s demand is given below:**

Y | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 | 11 | 12 |

P(Y) | \(\frac{ 1 }{ 36 }\) | \(\frac{ 2 }{ 36 }\) | \(\frac{ 3 }{ 36 }\) | \(\frac{ 4 }{ 36 }\) | \(\frac{ 5 }{ 36 }\) | \(\frac{ 6 }{ 36 }\) | \(\frac{ 5 }{ 36 }\) | \(\frac{ 4 }{ 36 }\) | \(\frac{ 3 }{ 36 }\) | \(\frac{ 2 }{ 36 }\) | \(\frac{ 1 }{ 36 }\) |

Therefore,

**Expectation of X = E(X) = \(\sum X_{ i }P\left (X_{ i } \right)\)**

= \(2 \times \frac{ 1 }{ 36 } + 3 \times \frac{ 2 }{ 36 } + 4 \times \frac{ 3 }{ 36 } + 5 \times \frac{ 4 }{ 36 } + 6 \times \frac{ 5 }{ 36 } + 7 \times \frac{ 6 }{ 36 } + 8 \times \frac{ 5 }{ 36 } + 9 \times \frac{ 4 }{ 36 } + 10 \times \frac{ 3 }{ 36 } + 11 \times \frac{ 2 }{ 36 } + 12 \times \frac{ 1 }{ 36 } \)

= \(\frac{ 2 }{ 36 } + \frac{ 6 }{ 36 } + \frac{ 12 }{ 36 } + \frac{ 20 }{ 36 } + \frac{ 30 }{ 36 } + \frac{ 42 }{ 36 } + \frac{ 40 }{ 36 } + \frac{ 36 }{ 36 } + \frac{ 30 }{ 36 } + \frac{ 22 }{ 36 } + \frac{ 12 }{ 36 }\)

= \(\frac{ 252 }{ 36 }\) **= 7**

**Expectation of X ^{2} = E(X^{2}) = \(\sum X_{ i }^{ 2 }P\left (X_{ i } \right)\)**

= \(4 \times \frac{ 1 }{ 36 } + 9 \times \frac{ 2 }{ 36 } + 16 \times \frac{ 3 }{ 36 } + 25 \times \frac{ 4 }{ 36 } + 36 \times \frac{ 5 }{ 36 } + 49 \times \frac{ 6 }{ 36 } + 64 \times \frac{ 5 }{ 36 } + 81 \times \frac{ 4 }{ 36 } + 100 \times \frac{ 3 }{ 36 } + 121 \times \frac{ 2 }{ 36 } + 144 \times \frac{ 1 }{ 36 } \)

= \(\frac{ 4 }{ 36 } + \frac{ 18 }{ 36 } + \frac{ 48 }{ 36 } + \frac{ 100 }{ 36 } + \frac{ 180 }{ 36 } + \frac{ 294 }{ 36 } + \frac{ 320 }{ 36 } + \frac{ 324 }{ 36 } + \frac{ 300 }{ 36 } + \frac{ 242 }{ 36 } + \frac{ 144 }{ 36 }\)

= \(\frac{ 1974 }{ 36 }\) **= 54.8333**

Then,

**Var(X) = E(X ^{2}) – [ E(X)^{2 }]**

= 54.8333 – 49 **= 5.8333**

Hence,

Standard derivation = \(\sqrt{Var\left (X \right)}\)

= \(\sqrt{ 5.8333 }\) **= 2.415**

** **

**Q.14: There is a class having 15 students with ages 14, 17, 15, 14, 21, 17, 19, 20, 16, 18, 20, 17, 16, 19 and 20 years. One student will be selected in such a way that each of them have the same chance of being chosen and Y is recorded as the age of the selected student. Find the probability distribution for the random variable Y. What will be the mean, variance and standard derivation of Y? **

** **

**Solution:**

A class has 15 students with ages 14, 17, 15, 14, 21, 17, 19, 20, 16, 18, 20, 17, 16, 19 and 20 years.

**Every student has the same chance of being selected. **

Hence, the probability for every student being selected is \(\frac{ 1 }{ 15 }\)

**The above information will be compiled in the frequency table which is given below:**

Y | 14 | 15 | 16 | 17 | 18 | 19 | 20 | 21 |

f | 2 | 1 | 2 | 3 | 1 | 2 | 3 | 1 |

**P(X = 14) =** \(\frac{ 2 }{ 15 }\)

**P(X = 15) =** \(\frac{ 1 }{ 15 }\)

**P(X = 16) =** \(\frac{ 2 }{ 15 }\)

**P(X = 17) =** \(\frac{ 3 }{ 15 }\)

**P(X = 18) =** \(\frac{ 1 }{ 15 }\)

**P(X = 19) =** \(\frac{ 2 }{ 15 }\)

**P(X = 20) =** \(\frac{ 3 }{ 15 }\)

**P(X = 21) =** \(\frac{ 1 }{ 15 }\)

**Hence, the probability distribution which is required as per the Q.’s demand is given below:**

Y | 14 | 15 | 16 | 17 | 18 | 19 | 20 | 21 |

P(Y) | \(\frac{ 2 }{ 15 }\) | \(\frac{ 1 }{ 15 }\) | \(\frac{ 2 }{ 15 }\) | \(\frac{ 3 }{ 15 }\) | \(\frac{ 1 }{ 15 }\) | \(\frac{ 2 }{ 15 }\) | \(\frac{ 3 }{ 15 }\) | \(\frac{ 1 }{ 15 }\) |

**Thus,**

**Expectation of X = E(X) = \(\sum X_{ i }P\left (X_{ i } \right)\)**

= \(14 \times \frac{ 2 }{ 15 } + 15 \times \frac{ 1 }{ 15 } + 16 \times \frac{ 2 }{ 15 } + 17 \times \frac{ 3 }{ 15 } + 18 \times \frac{ 1 }{ 15 } + 19 \times \frac{ 2 }{ 15 } + 20 \times \frac{ 3 }{ 15 } + 21 \times \frac{ 1 }{ 15 }\)

= \(\\\frac{ 28 }{ 15 } + \frac{ 15 }{ 15 } + \frac{ 32 }{ 15 } + \frac{ 51 }{ 15 } + \frac{ 18 }{ 15 } + \frac{ 38 }{ 15 } + \frac{ 60 }{ 15 } + \frac{ 21 }{ 15 }\)

= \(\frac{ 263 }{ 15 }\) **= 17.53**

**Expectation of X ^{2} = E(X^{2}) =** \(\sum X_{ i }^{ 2 }P\left (X_{ i } \right)\)

= \(\left (14 \right)^{ 2 } \times \frac{ 2 }{ 15 } + \left (15 \right)^{ 2 } \times \frac{ 1 }{ 15 } + \left (16 \right)^{ 2 } \times \frac{ 2 }{ 15 } + \left (17 \right)^{ 2 } \times \frac{ 3 }{ 15 } + \left (18 \right)^{ 2 } \times \frac{ 1 }{ 15 } + \left (19 \right)^{ 2 } \times \frac{ 2 }{ 15 } + \left (20 \right)^{ 2 } \times \frac{ 3 }{ 15 } + \left (21 \right)^{ 2 } \times \frac{ 1 }{ 15 }\)

= \(\frac{ 392 }{ 15 } + \frac{ 225 }{ 15 } + \frac{ 512 }{ 15 } + \frac{ 867 }{ 15 } + \frac{ 324 }{ 15 } + \frac{ 722 }{ 15 } + \frac{ 1200 }{ 15 } + \frac{ 441 }{ 15 }\)

= \(\frac{ 4683 }{ 15 }\) **= 312.2**

Then,

**Var(X) = E(X ^{2}) – [ E(X)^{2 }]**

= 312.2 – (17.53)^{2} **= 4.78**

Hence,

Standard derivation = \(\sqrt{Var\left (X \right)}\)

= \(\sqrt{ 4.78 }\) **= 2.186 **

** **

**Q.15: 70% members are in the favour and 40% are opposing a certain proposal in a meeting. A member will be randomly selected and we will take Y = 0, if the person opposed, and X = 1 if the person favours. What will be E(Y) and var(Y) in such a situation?**

** **

**Solution:**

As per the data given in the Q., we have

P(Y = 0) = 40% = \(\frac{ 40 }{ 100 }\) = 0.4

P(Y = 1) = (100 – 40)% = 60% = \(\frac{ 60 }{ 100 }\) **= 0.6**

**Hence, the probability distribution which is required as per the Q.’s demand is given below:**

X | 0 | 1 |

P(X) | 0.4 | 0.6 |

Thus,

**Expectation of X = E(X) =** \(\sum X_{ i }P\left (X_{ i } \right)\)

= 0 × 0.4 + 1 × 0.6 **= 0.6**

**Expectation of X ^{2} = E(X^{2}) =** \(\sum X_{ i }^{ 2 }P\left (X_{ i } \right)\)

= 0^{2} × 0.4 + 1^{2} × 0.6 **= 0.6**

We know that,

**Var(X) = E(X ^{2}) – [ E(X)^{2 }]**

= 0.6 – (0.6)^{2 }= 0.6 – 0.36 **= 0.24**

** **

**Q.16: The mean of the numbers which will be obtained after throwing a die on which 1 is written on two of the faces, 2 on three faces and 5 on one face is**

**(a) 1 **

**(b) 2**

** (c) 5**

**(d) \(\frac{ 13 }{ 6 }\)**

** **

**Solution:**

Let us consider Y which is a random variable representing any number on the die.

Since, it’s a die, then the total number of observations will be six.

Thus, P(Y = 1) = \(\frac{ 2 }{ 6 } = \frac{ 1 }{ 3 }\)

P(Y = 2) = \(\frac{ 3 }{ 6 } = \frac{ 1 }{ 2 }\)

P(Y = 5) = \(\frac{ 1 }{ 6 }\)

**Hence, the probability distribution which is required as per the Q.’s demand is given below:**

Y | 1 | 2 | 5 |

P(Y) | \(\frac{ 1 }{ 3 }\) | \(\frac{ 1 }{ 2 }\) | \(\frac{ 1 }{ 6 }\) |

Thus,

Expectation of X = E(X) = \(\sum X_{ i }P\left (X_{ i } \right)\)

= \(1 \times \frac{ 1 }{ 3 } + 2 \times \frac{ 1 }{ 2 } + 5 \times \frac{ 1 }{ 6 }\)

= \(\frac{ 1 }{ 3 } + 1 + \frac{ 5 }{ 6 }\) = \(\frac{ 2 + 6 + 5 }{ 6 }\) = \(\frac{ 13 }{ 6 }\)

**Hence, the correct answer is (d).**

** **

**Q.17: Let Y be the number of aces obtained. Let any of the two cards are drawn at random from the deck of the cards. Then, the value of E(X) will be**

**(a) \(\frac{ 37 }{ 221 }\) **

**(b) \(\frac{ 5 }{ 13 }\)**

**(c) \(\frac{ 2 }{ 13 }\)**

**(d) \(\frac{ 1 }{ 13 }\)**

** **

**Solution:**

Let, Y be the number of aces obtained.

Therefore, Y will take any of the values of 0, 1, or 2

As we know, there are 52 cards in a deck. Among them 4 cards are aces.

Hence, there are 48 non- ace cards.

**P(Y = 0) = P(2 non- ace cards and 0 ace card) = **\(\frac{ ^{ 48 }C_{ 2 } \times ^{ 4 }C_{ 0 } }{ ^{ 52 }C_{ 2 } } = \frac{ 1128 }{ 1326 }\)

**P(Y = 1) = P(1 non- ace cards and 1 ace card) =** \(\frac{ ^{ 48 }C_{ 1 } \times ^{ 4 }C_{ 1 } }{ ^{ 52 }C_{ 2 } } = \frac{ 192 }{ 1326 }\)

**P(Y = 2) = P(0 non- ace cards and 2 ace card) =** \(\frac{ ^{ 48 }C_{ 0 } \times ^{ 4 }C_{ 2 } }{ ^{ 52 }C_{ 2 } } = \frac{ 6 }{ 1326 }\)

**Hence, the probability distribution which is required as per the Q.’s demand is given below:**

Y | 0 | 1 | 2 |

P(Y) | \(\frac{ 1128 }{ 1326 }\) | \(\frac{ 192 }{ 1326 }\) | \(\frac{ 6 }{ 1326 }\) |

Thus,

Expectation of X = E(X) = \(\sum X_{ i }P\left (X_{ i } \right)\)

= \(0 \times \frac{ 1128 }{ 1326 } + 1 \times \frac{ 192 }{ 1326 } + 2 \times \frac{ 6 }{ 1326 }\)

= \(\frac{ 192 }{ 1326 } + \frac{ 12 }{ 1326 }\) = \(\frac{ 192 + 12 }{ 1326 }\) = \(\frac{ 204 }{ 1326 }\) = \(\frac{ 2 }{ 13 }\)

**Hence, the correct answer is (c).**