Ncert Solutions For Class 12 Maths Ex 13.2

Ncert Solutions For Class 12 Maths Chapter 13 Ex 13.2

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.

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

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

= 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,

\(P (Q\cap R) = P (Q)\cdot P(R) = (0.3) (0.4) = 0.12\)

 

(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:

\(P (Q’ \cup R’) = \frac{1}{4}\)

= \( P ((Q \cap R)’) = \frac{1}{4}\)

= 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).

 

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