**Q.1: If P (Q) = \(\frac{3}{5}\) & P (R) = \(\frac{1}{5}\) . Considering events Q and R are independent.**

** **

**Sol:**

Given, P (Q) = \(\frac{3}{5}\) & P (R) = \(\frac{1}{5}\)

Q and R are the events which are independent in nature.

\( P (Q \ cap R) = P (Q) \times P(R) \)= \( \frac{3}{5} \times \frac{1}{5} \) = \( \frac{3}{25} \)

**Q.2: 2 cards are drawn at random from a well shuffled deck of 52 cards. Determine the probability that the drawn cards are black.**

** **

**Sol:**

In a well shuffled deck of 52 cards there are only 26 black cards (13 spade cards and 13 clubs cards)

Let Q be the event black card is drawn from a deck of cards.

P (Q) = \(\frac{26}{52} = \frac{1}{2}\)

Let R be the probability a black card on the second draw.

P (R) = \(\frac{25}{52}\)

**Thus, the probability of getting both black cards =** (\(\frac{26}{52}\))(\(\frac{25}{52}\)) = \(\frac{25}{102}\)

**Q.3: A bag full of oranges is inspected by selecting 3 random oranges without replacement. If it is seen that all the 3 drawn oranges are good then the box is sent for sale, otherwise it is sent for rejection. Determine the probability if a box containing 15 oranges out of which 12 of which are good and rest are bad will be approved for sale or not? **

** **

**Sol:**

Let us consider events Q, R and S be the events of drawing first, second and third drawing of orange.

**P (Q) be the event of first drawing of orange =** \(\frac {12} {15} \)

**P(R) be the event of second drawing of orange =** \(\frac {11} {14} \)

**P(S) be the event of third drawing of orange =** \(\frac {10} {13} \)

**Thus , getting the probability of all good apples =** (\(\frac{12}{15}\))(\(\frac{11}{14}\))(\(\frac{10}{13}\))

= \(\frac {44} {91} \)

Thus, the probability of getting approved for sale is \(\frac {44} {91} \)

**Q.4: An unbiased coin is tossed. An EVENT Q and R be the head appears on the coin & 3 appears on the face of the die. Find out whether Q and R are independent events or not?**

** **

**Sol:**

**Sample space = { (T, 6), (T, 5), (T, 4), (T, 3), (T, 2), (T, 1), (H, 6), (H, 5), (H, 4), (H, 3), (H, 2), (H, 1) }**

Let Q be the event of head appearing on the face of the coin

**Q = { (H , 6) , (H, 5) , (H , 4) , (H , 3) , (H , 2) , (H , 1) }**

P (Q) = \( \frac{6}{12} = \frac{1}{2} \)

Let R be the event that the number 3 appears on the face of the die

**R = {(T, 3), (H, 3)}**

P (R) = \( \frac{2}{12} = \frac{1}{6} \)

Therefore,

\(Q \cap R \) = {(H, 3)}

P (\( Q \cap R \)) = \(\frac {1}{12} \)

\(P (Q) \times P(R) \) = \( \frac{1}{2} \times \frac {1}{6} \) = \( P (Q) \cap P(R) \)

**Therefore, Q & R are independent events.**

**Q.5: An experiment consists of marking a die 1, 2, 3 in red and 4, 5, 6 in green and then rolled. Let Q be the event that the number is even and R be the event that color of the face of the die is red. Determine the probability that both of the events are independent.**

** **

**Sol:**

Sample space (S) when a die is rolled = {6, 5, 4, 3, 2, 1}

According to the Q., it is given that:

Q = Event that the number appears is even

R = the colour of the face of the die is red.

Q = {6, 4, 2}

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

R = {3, 2, 1}

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

Therefore, \( Q \cap R = {2} \)

\(P (QR) = P (Q \cap R) = \frac{1}{6}\) \(P (Q) \cdot P (R) = \frac{1}{2} \times \frac{1}{2} = \frac{1}{4} ≠ \frac{1}{6}\)**Hence, events Q and R are not independent events. **

** **

**Q.6: Q and R be the events such that P (Q) = \(\frac{3}{5}\) , P (R) = \(\frac{3}{10}\) & \(Q \cap R =\frac{1}{5}\) . Determine whether events Q and R are independent or not /**

** **

**Sol:**

According to the Q., it is given that:

P (Q) = \(\frac{3}{5}\) , P (R) = \(\frac{3}{10}\) & \(Q \cap R =\frac{1}{5}\) .

\(P (Q) \cdot P (R) = \frac{3}{5} \times \frac{3}{10} = \frac{9}{50} ≠ \frac{1}{5}\)**Hence, events Q and R are not independent events. **

**Q.7: Q and R be the events such that P (Q) = \(\frac{1}{2}\) , \(Q \cap R =\frac{3}{5}\) , P (R) = p . Determine the value of p if Q and R are:**

**(a) Mutually exclusive **

**(b) Independent.**

** **

**Sol:**

**(a) Let us consider Q and R are mutually exclusive. **

\(Q \cap R =\phi \), P (R) = p

\(P (Q \cap R) = 0\)We know, that:

\(P (Q \cap R) = P (Q) + P(R) – P (Q \cup R)\)= \(\frac {3} {5} = \frac{1}{2} + p – 0\)

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

**(b) Considering events Q and R are independent events. **

= \(\frac{3}{5} = \frac{1}{2} + p – \frac{1}{2}p\)

= p = \(\frac{2}{10} = \frac{1}{5}\)

**Q.8: Let Q and R be the independent events such that P (Q) = 0.3, P (R) = 0.4**

**Determine:**

**(a) P (\( P (Q \cap R) \)) **

**(b) P (\( P (Q \cup R) \))**

**(c) \(P (\frac{Q}{R})\)**

**(d) \(P (\frac{R}{Q})\)**

** **

**Sol:**

**(a) Considering the events Q and R independent events, **

**(b) \( P(Q \cap R) = P(Q) + P(R) – P (Q \cup R) \)**

P (\( P (Q \cup R) \)) = 0.3 + 0.4 – 0.12 **= 0.58**

**(c) \(P (\frac{Q}{R}) = \frac{P(Q \cap R)}{P(R)}\)**

= \( P (\frac{Q}{R}) = \frac{0.12}{0.4}\) **= 0.3**

**(d) \(P (\frac{R}{Q}) = \frac{P(Q \cap R)}{P(Q)}\)**

= \( P (\frac{R}{Q}) = \frac{0.12}{0.3}\) **= 0.4**

**Q.9: Let Q and R be the two events such that \( P (Q) = \frac{1}{2}\), \( P (R) = \frac{1}{2}\), \( P (Q \cap R) = \frac {1} {8} \). Determine the value of P (neither Q nor R)**

** **

**Sol:**

According to the Q., it is given that:

\(P (Q) = \frac{1}{2}\), \( P (Q \cap R) = \frac{1}{8}\), \( P (R) = \frac{1}{2}\)

P (neither Q or R) = P \( (Q’ \cap R’) \)

P (neither Q or R) = P \( (Q \cap R)’ \)

= 1 – \( P (Q \cap R) \)

= 1 – \( P (Q) + P(R) – P (Q \cap R)\)

= 1 – \(\frac{1}{4} + \frac{1}{2} – \frac{1}{8}\)

= 1 – \( \frac{5}{8}\) = \( \frac{3}{8} \)

**Q.10: Q and R are events such that P(Q) = \( \frac {1}{2}\) , P(R) = \( \frac {1}{2}\) & P(neither Q nor R) = \( \frac {1}{4}\)**

**Sol:**

= \( P ((Q \cap R)’) = \frac{1}{4}\)

= 1 – \( P (Q \cap R) = \frac{1}{4}\)

= \( P (Q \cap R) = \frac{3}{4}\)

= \( P (Q\cap R) = P (Q)\cdot P(R) = (\frac {1} {2}) (\frac{7}{12}) = \frac{7}{24}\)

\( \frac{3}{4} ≠ \frac{7}{24}\)**Hence, Q & R are not independent events.**

**Q.11: Q and R are two independent events such that P (Q) 0.3 & P(R) = 0.6. Determine:**

**(a) P (Q and R)**

**(b) P (Q and not R) **

**(c) P (Q or R) **

**(d) P (neither Q nor R) **

** **

**Sol:**

**(a) P (Q and R) =** \( P (Q\cap R) = P (Q)\cdot P(R) = (0.3) (0.6) = 0.18\)

**(b) P (Q and not R) **

= \( P (Q \cap R’) \)

= \( P (Q) – P (Q \cap R) \)

= 0.3 – 0.18 **= 0.12**

**(c) P (Q or R) **

= \( (Q \cap R) \)

= \( P (Q) + P(R) – P (Q \cap R) \)

= 0.3 + 0.6 – 0.18 **= 0.72**

**(d) P (neither Q nor R) **

= \( P (Q’ \cap R’) \)

= \( P ((Q \cap R)’)\)

= 1 – \( P (Q \cap R) \)

= 1 – 0.72 **= 0.28**

**Q.12: An experiment consists of tossing up of die thrice. Determine the probability of getting an odd number at least once.**

** **

**Sol:**

Probability of getting an odd number in a single throw = \(\frac{3}{6} = \frac{1}{2}\)

Probability of getting an even number = \(\frac {3} {6} = \frac{1}{2}\)

Probability of getting an even number = \(\frac {1} {2}\times \frac{1}{2}\times \frac{1}{2}\) = \(\frac {1} {8} \)

According to the Q.,

Determining the probability of getting an odd number at least once

= 1 – Probability of getting an odd number

= 1 – probability of getting an even number thrice

= 1 – \(\frac{1}{8}\) = \(\frac{7}{8}\)

**Q.13: From a bag of 10 black balls and 8 red balls 2 balls are drawn at a random without random. Determine the probability:**

**(a) Both the drawn balls are red**

**(b) First and the second drawn ball are black & red respectively.**

**(c) One of them is black and the other one is red.**

** **

**Sol:**

According to the Q., it is given that:

The bag contains a total number of number balls of = 18

**No. of red balls = 8**

**No. of black balls = 10**

**(a) Probability that a red ball appears in the first draw =** \(\frac {8} {18} = \frac{4}{19}\)

The ball so obtained is replaced.

**Probability that a red ball appears in the second draw =** \(\frac{8}{18} = \frac{4}{19}\)

**Probability that both the red balls appear =** \(\frac{4}{9} \times\frac{4}{9} = \frac{16}{81}\)

**(b) Probability that a black ball appears in the first draw =** \(\frac{10}{18} = \frac{5}{19}\)

**The ball so obtained is replaced.**

**Probability that a black ball appears in the second draw** = \(\frac {8} {18} = \frac{4}{19}\)

**Probability that both the black balls appear =** \(\frac {5}{9} \times\frac{4}{9} = \frac{20}{81}\)

** **

**(c) Probability that a red ball appears in the first draw =** \(\frac {8} {18} = \frac{4}{19}\)

The ball so obtained is replaced.

**Probability that a black ball appears in the second draw =** \(\frac {8} {18} = \frac{5}{9}\)

**Probability that both the red balls appear =** \(\frac {4} {9} \times\frac{5}{9} = \frac{20}{81}\)

**Hence, the probability of getting one black and one red ball =**

**Probability of getting first red ball and then black ball + Probability of getting first black ball and then red ball **

= \(\frac {20} {81} + \frac{20}{81}\) = \(\frac {40} {81} \)

**Q.14: The probability of solving particular problems by methods Q and R are \(\frac{1}{2}\) and \(\frac{1}{3}\) respectively. Considering that the problem is solved independently, determine the probability:**

**(a) The problem is solved **

**(b) At least one of them solves the problem correctly.**

** **

**Sol:**

Probability that the problem is solved by Q, P (Q) = \(\frac {1} {2} \)

Probability that the problem is solved by R, P (R) = \(\frac {1} {3} \)

According to the Q., it is given that both problems are solved independently by Q and R:

= P (QR) = \( P (Q) \times P(R) = \frac {1} {2} \times \frac{1}{3} = \frac {1} {6} \)

**P (Q’) = 1 – P (Q) =** 1 – \( \frac {1} {2} = \frac {1} {2} \)

**P (R’) = 1 – P(R) =** 1 – \( \frac {1} {3} = \frac {2} {3} \)

**(a) Probability that the problem:**

= \( P (Q \cap R) \)

= P (Q) + P(R) – P (QR)

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

**(b) Probability that exactly one solves the above mentioned problem:**

= \( P (A) \times P (B’) + P (B) \times P (A’) \)

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

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

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

**Q.15: An experiment consists of drawing up of a card at random from a well shuffled deck of 52 cards. Events Q and R associated with the following experiment are independent in nature **

**(a) Q: the drawn card is a spade**

** R: the drawn card is an ace**

**(b) Q: it is a black card**

** R: the drawn card is a king**

**(c) Q: the drawn card can either be a king or queen **

** R: the drawn card is either a queen or jack.**

**Sol:**

** (a) Total number of spades = 13**

**Total number of ace cards in a pack of cards = 4**

**Probability of drawing a spade card P (Q) =** \( \frac{13}{52} = \frac{1}{4}\)

**Probability of drawing an ace card P(F) =** \( \frac{1}{4}\)

**In a pack of cards, there is only card which is an ace of spades. Hence probability of getting an ace of spades =** \(\frac{1}{52}\)

= \( P (Q) \times P(R) \)

= \( \frac{1}{4} \times \frac{1}{13} \) = P (QR)

P (QR) = \( P (Q) \times P(R) \)

**Hence, events Q and R are independent in nature.**

** **

**(b) In a pack of 52 cards, there are 26 black cards and 4 cards which are kings.**

**P (Q) = P (the drawn card is black) =** \(\frac{26}{52} = \frac{1}{2}\)

**P (R) = P (the drawn card is a king) =** \(\frac {4}{52} = \frac{1}{13}\)

In a pack of 52 cards, there are 2 black cards which are kings.

P (QR) = P (the drawn card is black in colour as well as a king) = \(\frac{2}{52} = \frac{1}{26}\)

= \( P(Q) \times P(R) \)

= \( \frac{1}{2} \times \frac{1}{13} = \frac{1}{26}\) = P (QR)

**Hence, events Q and R are independent in nature.**

** **

**(c) In a pack of 52 cards, there are 4 kings, 4 jacks and 4 queens.**

P (Q) = P (the drawn card can either be a king or queen) = \(\frac{8}{52} = \frac{2}{13}\)

P (R) = P (the drawn card is a queen or a jack) = \(\frac{8}{52} = \frac{2}{13}\)

**There are 4 cards which are king or queen and queen or jack =** \(\frac{4}{52} = \frac{1}{13}\)

= \( P(Q) \times P(R) \)

= \( \frac{2}{13} \times \frac{2}{13} ≠ \frac{4}{169}\) = P (QR)

**Hence, events Q and R are not independent in nature.**

**Q.16: Among the students residing in a hostel, 60% of them read Hindi newspaper, 20% of them read both Hindi and English newspapers. A student is selected as a random:**

**(a) Determine the probability the students reads neither Hindi nor English newspapers.**

**(b) Determine the probability that the student reads Hindi newspapers, but also reads English newspaper too.**

**(c) If the students read English newspapers, find the probability that they reads Hindi newspaper too.**

** **

**Sol:**

**According to the Q. it is given that Q be the event who reads Hindi newspapers and R be the event who reads English newspapers.**

P (Q) = 60% = \(\frac{60}{100} = \frac{3}{5}\)

P (R) = 40% = \(\frac{40}{100} = \frac{2}{5}\)

P (\( Q \cap R \)) = 20% = \(\frac{20}{100} = \frac{1}{5}\)

**(a) Probability that the student reads Hindi or English newspapers:**

= \((Q \cup R)’\)

= 1 – \((Q \cup R)\)

= 1 – { P(Q) + P(R) – \((Q \cup R)\) }

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

= 1 – \(\frac{4}{5}\) = \(\frac{1}{5}\)

**(b) Probability that a guy chose to read English newspaper , if they already reads Hindi newspaper =** \(P (\frac{Q}{R})\)

= \(\frac{Q \cap R}{R}\) = \(\frac{\frac{1}{5}}{\frac{3}{5}}\) = \(\frac{1}{3}\)

**(c) Probability that a guy chose to read Hindi newspaper, if they already reads English newspaper = \(P (\frac{R}{Q})\)**

= \(\frac{Q \cap R}{Q}\) = \(\frac{\frac{1}{5}}{\frac{2}{5}}\) = \(\frac{1}{2}\)

**Q.17: Find the probability of obtaining an even number, when two dice are rolled at once:**

**(i) 0**

**(ii) \(\frac{1}{3}\)**

**(iii) \(\frac{1}{12}\)**

**(iv) \(\frac{1}{36}\)**

** **

**Sol:**

Total number of outcomes when two dice are rolled at once = 36

2 is the only even prime number.

Let Q be the event of getting an even prime number.

Therefore, Q = {2, 2}

P (Q) = \(\frac{1}{36}\)

**Hence, the correct answer to the above mentioned Q. is (iv)**

**Q.18: 2 events Q and R will be independent, such as:**

**(i) Q and R are mutually exclusive.**

**(ii) P (Q’R’) = [1 – P (Q) ][1 – P(R) ]**

**(iii) P(Q) = P (R)**

**(iv) P (Q) + P (R) = 1**

** **

**Sol:**

2 events Q and R are said to be independent, if

P (QR) = \( P (Q) \times P (R) \)

= P (Q’R’) = [ 1 – P(Q)][1 – P(R)]

= \(P(Q’ \cap R’)\) = \( 1 – P(Q) – P(R) + P(Q) \cdot P(R) \)

= 1- \(P(Q’ \cap R’)\) = \( P(Q) – P(R) + P(Q) \cdot P(R) \)

= \(P(Q \cup R)\) = \( P(Q) – P(R) + P(Q) \cdot P(R) \)

= P (Q) + P (R) – P (QR) = \( P(Q) – P(R) + P(Q) \cdot P(R) \)

= P (QR) = \( P(R) + P(Q) \cdot P(R) \)

**This implies that event Q and R are independent event, **

P (Q’R’) = [1 – P(Q)][1 – P(R)]

Let P (Q) = a

P (R) = b

0 ˂ a, b ˂ 1

Q and R are both mutually exclusive events.

Therefore,

= \(P(Q \cup R) = \phi \)

** = P (QR) = 0**

Therefore, P(Q) × P (R) = ab ≠ 0

Therefore, \( P (Q) \times P (R) ≠ P (QR) \)

** (ii) Event of getting an even number = {6, 4, 2}**

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

= \(P(Q \cup R) = \phi \)

= \( P (Q) \times P (R) = \frac {1}{4} ≠ 0 \)

**(iii) Let Q be the event of getting a odd number = {5 , 3, 1 }**

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

**(iv) P(Q) + P(R) =** \(\frac{1}{2}\) + \(\frac{1}{2}\) = 1

**From the above solution it cannot be concluded that Q and R are independent events.**

**Therefore, the correct answer to the above mentioned Q. is (ii).**