Referring to the AP Inter 2nd Year Maths Study Material Chapter 13 Probability Exercise 13b Solutions makes it easier to understand complex problems.
AP Inter 2nd Year Maths Probability Solutions Exercise 13b
I.
Question 1.
If P(A) = \(\frac{3}{5}\) and P(B) = \(\frac{1}{5}\), find P (A ∩ B) if A and B are independent events.
Solution:
Given that A and B are independent events P(A) = \(\frac{3}{5}\) and P(B) = \(\frac{1}{5}\)
∴ P(A ∩ B) = P(A) P(B) = \(\frac{3}{5}\) × \(\frac{1}{5}\) = \(\frac{3}{25}\)
Question 2.
Two cards are drawn at random in succession and without replacement from a pack of 52 playing cards. Find the probability that both the cards are black.
Solution:
There are 26 black cards in a deck of 52 cards.
Let A,B be the events of drawing black cards in the successive draws.
The probability of getting a black card in the first draw is P(A) = \(\frac{26}{52}\)
The probability of getting a black card in the second draw is P(B|A) = \(\frac{25}{51}\)
Since the drawn card is not replaced, A and B are dependent events.
∴ Probability of getting both the cards black P(A ∩ B) = P(A)P(B|A) = \(\frac{26}{52}\) × \(\frac{25}{52}\) = \(\frac{25}{102}\)
Question 3.
Let E and F be events with P(E) = \(\frac{3}{5}\), P(F) = \(\frac{3}{10}\) and P(E ∩ F) = \(\frac{1}{5}\). Are E and F independent?
Solution:
Given P(E) = \(\frac{3}{5}\), P(F) = \(\frac{3}{10}\) and P(E ∩ F) = \(\frac{1}{5}\)
Now P(E) P(F) = \(\frac{3}{5}\) × \(\frac{3}{10}\) = \(\frac{9}{50}\) ≠ \(\frac{1}{5}\)
P(E)P(F) ≠ P(E ∩ F)
∴ E and F are not independent.
II.
Question 1.
A box of oranges is inspected by examining three randomly selected oranges drawn without replacement. If all the three oranges are good, the box is approved for sale, otherwise, it is rejected. Find the probability that a box containing 15 oranges out of which 12 are good and 3 are bad ones will be approved for sale.
Solution:
Let A, B, C be the respective events that the first, second, and the third drawn orange is good.
Probability that first drawn orange is good, P(A) = \(\frac{12}{13}\)
The oranges are not replaced.
Probability of getting second orange is good, P (B) = \(\frac{11}{14}\)
Probability of getting third orange is good, P(C) = \(\frac{10}{13}\)
Since the box is approved for sale, if all the three oranges are good.
∴ Probability of getting all the oranges good
P(A ∩ B ∩ C) = P(A)P(B)P(C) = \(\frac{12}{15}\) × \(\frac{11}{14}\) × \(\frac{10}{13}\) = \(\frac{44}{91}\)
Question 2.
A fair coin and an unbiased die are tossed. Let A be the event head appears on the coin’ and B be the event 43 on the die’. Check whether A and B are independent events or not.
Solution:
The sample space is given by,
S = {(H, 1), (H, 2) (H, 3), (H, 4), (H, 5), (H, 6)
(T, 1),(T, 2),(T, 3),(T, 4),(T, 5),(T, 6)}
Let A: Head appears on the coin
⇒ P(A) = \(\frac{6}{12}\) = \(\frac{1}{2}\)
A = {(H, 1), (H, 2), (H, 3), (H, 4), (H, 5), (H, 6)}
B: 3 on the die ⇒ B = {(H, 3),(T, 3)} ⇒ P(B) = \(\frac{2}{12}\) = \(\frac{1}{6}\)
Now, A ∩ B = {(H, 3)} ⇒ P(A ∩ B) = \(\frac{1}{12}\)
Hence, P(A)P(B) = \(\frac{1}{2}\) × \(\frac{2}{6}\) = P(A ∩ B)
∴ A and B are independent events.
Question 3.
A die marked 1, 2, 3 In red and 4, 5, 6 in green is tossed. Let A be the event, ‘the number ¡s even,’ and B be the event, ‘the number Is red’. Are A and B independent?
Solution:
When a die is tossed, the sample space is S = {1, 2, 3, 4, 5, 6}
Here, 1, 2, 3 are red in colour and 4, 5, 6 are green.
Let A: the number is even {2, 4, 6} ⇒ P(A) = \(\frac{3}{6}\) = \(\frac{1}{2}\)
B: the number is red = {1, 2, 3} ⇒ P(B) = \(\frac{3}{6}\) = \(\frac{1}{2}\)
∴ A ∩ B = {2}
P(A ∩ B) = \(\frac{1}{6}\)
Now P(A) P(B) = \(\frac{1}{2}\) × \(\frac{1}{2}\) = \(\frac{1}{4}\) ≠ \(\frac{1}{6}\)
∴ P(A)P(B) ≠ P(AB)
∴ A and B are not independent events.
Question 4.
Given that the events A and B are such that P(A) = \(\frac{1}{2}\), P(A ∪ B) = \(\frac{3}{5}\) and P(B) = p, Find p if they are (i) mutually exclusive (ii) independent.
Solution:
Given that P(A) = \(\frac{1}{2}\), P(A ∪ B) = \(\frac{3}{5}\) and P(B) = p
(i) When A and B are mutually exclusive,, A ∩ B = Φ
∴ P(A ∩ B) = 0
We know that P(A ∪ B) = P(A) + P(B) – P(A ∩ B)
⇒ \(\frac{3}{5}\) = \(\frac{1}{2}\) + p = 0
⇒ p = \(\frac{3}{5}\) – \(\frac{1}{2}\) = \(\frac{1}{10}\)
(ii) When A and B are independent, P(A ∩ B) = P (A) P(B) = \(\frac{1}{2}\)p
We know that P(A ∪ B) = P(A)+P(B)-P(A ∩ B)
⇒ \(\frac{3}{5}\) = \(\frac{1}{2}\) + p – \(\frac{1}{2}\)p
⇒ \(\frac{3}{5}\) = \(\frac{1}{2}\) + \(\frac{p}{2}\)
⇒ \(\frac{p}{2}\) = \(\frac{3}{5}\) – \(\frac{1}{2}\) = \(\frac{1}{10}\)
⇒ \(\frac{p}{2}\) = \(\frac{2}{10}\) = \(\frac{1}{5}\)
Question 5.
Let A and B be independent events with P(A) = 0.3 and P(B) = 0.4. Find
(i) P(A ∩ B)
(ii) P(A ∪ B)
(iii) P (A|B)
(iv) P (B|A)
Solution:
Given P (A) = 0.3 and P(B) = 0.4.
(i) If A and B are independent events, then P(A ∩ B) = P(A) × P(B) = 0.3 × 0.4 = 0.12
(ii) P(A ∪ B) = P(A) + P(B) – P (A ∩ B) ⇒ P(A ∪ B) = 0.3 + 0.4 – 0.12 = 0.58
(iii) P(A|B) = \(\frac{P(A \cap B)}{P(B)}=\frac{0.12}{0.4}\) = 0.3
(iv) P(B|A) = \(\frac{P(A \cap B)}{P(A)}=\frac{0.12}{0.3}\) = 0.4
Question 6.
If A and B are also events such that P(A) = \(\frac{1}{4}\), P(B) = \(\frac{1}{2}\) and P(A ∩ B) = \(\frac{1}{8}\), find P(not A and not B).
Solution:
Given that P(A) = \(\frac{1}{4}\), P(B) = \(\frac{1}{2}\) and P(A ∩ B) = \(\frac{1}{8}\)
P(not on Aand not on B) = P(A’ ∩ B’)
P(not on A and not on B) = P[(A ∪ B)’] [∵ A’ ∩ B = (A ∪ B)’]
= 1 – P(A ∪ B) = 1 – [P(A) + P(B) – P(AB)]
= 1 – [\(\frac{1}{4}\) + \(\frac{1}{2}\) – \(\frac{1}{8}\)] = 1 – \(\frac{5}{8}\)
= \(\frac{3}{8}\)
Question 7.
Events A and B are such that P(A) = \(\frac{1}{2}\), P(B) = \(\frac{7}{12}\) and P(not A or not B) = \(\frac{1}{4}\). State whether A and B are independent ?
Solution:
Given that P(A) = \(\frac{1}{2}\),P(B) = \(\frac{7}{12}\) and P (not A or not B) = \(\frac{1}{4}\),
⇒ P(A’ ∪ B’) = \(\frac{1}{4}\) ⇒ P[(A ∩ B)’] = \(\frac{1}{4}\) [∵ A’ ∪ B’ = (A ∩ B)’]
⇒ 1 – P(A ∩ B) = \(\frac{1}{4}\) ⇒ P(A ∩ B) = \(\frac{3}{4}\) …………. (1)
But P(A)P(B) = \(\frac{1}{2}\) × \(\frac{7}{12}\) = \(\frac{7}{24}\) ………….. (2)
Here, \(\frac{3}{4}\) ≠ \(\frac{7}{24}\)
∴ P(A ∩ B) ≠ P(A)P(B)
∴ A and B are not independent events.
Question 8.
Given two independent events A and B such that P(A) = 0.3, P(B) = 0.6. Find
(i) P(A and B)
(ii) P(A and not B)
(iii) P(A or B)
(iv) P(neither A nor B)
Solution:
Given that P(A) = 0.3, P(B) = 0.6 are independent events.
(i) P(A and B) = P(A ∩ B)=P(A)P(B) = 0.3 × 0.6 = 0.18
(ii) P(A and not B) = P(A ∩ B) = P(A)’ – P(A ∩ B) = 0.3 – 0.18 = 0.12
(iii) P(A or B) = P(A ∪ B)=P(A)+ P (B)- P(A ∩ B)=0.3+0.6 – 0.18 = 0.72
(iv) P (neither A nor B) = P(A ∩ B ) = P [(A ∪ B)’] [∵ A’ ∩ B’ =(A ∪ B)’]
= 1 – P(A ∪ B) = 1 – 0.72 = 0.28
Question 9.
A die is tossed thrice. Find the probability of getting an odd number at least once.
Solution:
Probability of getting an odd number in a single throw of a die = \(\frac{3}{6}\) = \(\frac{1}{2}\)
Probability of getting an even number = \(\frac{3}{6}\) = \(\frac{1}{2}\)
Probability of getting an even number three times = \(\frac{1}{2}\) × \(\frac{1}{2}\) × \(\frac{1}{2}\) = \(\frac{1}{8}\)
= 1 – Probability of getting an odd number in none of the throws
= 1 – Probability of getting an even number thrice = 1 – \(\frac{1}{8}\) = \(\frac{7}{8}\)
Question 10.
Two balls are drawn at random with replacement from a box containing 10 black and 8 red balls. Find the probability that
(i) both balls are red.
(ii) first ball is black and second is red.
(iii) one of them is black and other is red.
Solution:
Given that total number of balls = 18,
Number of red balls = 8,
Number of black balls = 10
(i) Probability of getting a red ball in the first draw = \(\frac{8}{18}\) = \(\frac{4}{9}\) . The ball is replaced after the first draw.
So, probability of getting a red ball in the second draw = \(\frac{8}{18}\) = \(\frac{4}{9}\)
∴ Probability of getting both the balls red = \(\frac{4}{9}\) × \(\frac{4}{9}\) = \(\frac{16}{81}\)
(ii) Probability of getting first ball black = \(\frac{10}{18}\) = \(\frac{5}{9}\).The ball is replaced after the first draw.
So, Probability of getting second ball as red = \(\frac{8}{18}\) = \(\frac{4}{9}\)
∴ Probability of getting first ball as black and second ball as red = \(\frac{5}{9}\) × \(\frac{4}{9}\) = \(\frac{20}{81}\)
(iii) Probability of getting first ball as red = \(\frac{8}{18}\) = \(\frac{4}{9}\).The ball is replaced after the first draw.
So, Probability of getting second ball as black = \(\frac{10}{18}\) = \(\frac{4}{9}\)
∴ Probability of getting first ball as black and second ball as red = \(\frac{4}{9}\) × \(\frac{5}{9}\) = \(\frac{20}{81}\)
Now probability that one of them is black and other is red = Probability of getting first ball black and second red + Probability of getting first ball red and second ball black
= \(\frac{20}{81}\)+ \(\frac{20}{81}\) = \(\frac{40}{81}\)
Question 11.
Probability of solving specific problem independently by A and B are \(\frac{1}{2}\) and \(\frac{1}{3}\) respectively. If both try to solve the problem independently, find the probability that (i) the problem is solved(ii) exactly one of them solves the problem.
Solution:
Probability of solving the problem by A is P(A) = \(\frac{1}{2}\)
Probability of solving the problem by B is P(B) = \(\frac{1}{3}\)
Since the problem is solved independently by A and B, we have
P(AB) = P(A) P(B) = \(\frac{1}{2}\) × \(\frac{1}{3}\) = \(\frac{1}{6}\)
Now P(A’) = 1 – P(A) = 1 – \(\frac{1}{2}\) = \(\frac{1}{2}\);
P(B’) = 1 – P(B) = 1 – \(\frac{1}{3}\) = \(\frac{2}{3}\)
(i) Probability that the problem is solved is
P(A ∪ B) = P(A) – P(B) – P(AB) = \(\frac{1}{2}\) + \(\frac{1}{3}\) – \(\frac{1}{6}\) = \(\frac{4}{6}\) = \(\frac{2}{3}\)
(ii) Probability that exactly one of them solves the problem
= P(A)P(B’) + P(B)P(A’) = \(\frac{1}{2}\) × \(\frac{2}{3}\) + \(\frac{1}{2}\) × \(\frac{1}{3}\) = \(\frac{1}{6}\) + \(\frac{1}{6}\) = \(\frac{1}{2}\)
III.
One card is drawn at random from a well shuffled deck of 52 cards. In which of the following cases are the events E and F independent ?
(i) E: ‘the card drawn ¡s a spade’
F: ‘the card drawn is an ace’
(ii) F : ‘the card drawn is black’
F: ‘the card drawn Is a king’ .
(iii) E : ‘the card drawn is a king or queen’
F : ‘the card drawn is a queen or jack’.
Solution:
(i) In a deck of 52 cards, 13 cards are spades and 4 cards are aces.
∴ P(E) = P (the card drawn is a spade) =\(\frac{13}{52}\) = \(\frac{1}{4}\)
Also P(F) = P(the card drawn in an ace) = \(\frac{4}{52}\) = \(\frac{1}{13}\)
In the deck of cards, an ace of spades is only one.
P(EF) = P (the card drawn is spade and an ace) = \(\frac{1}{52}\)
Now P(E) P(F) \(\frac{1}{4}\) × \(\frac{1}{13}\) = \(\frac{1}{52}\) ≠ P(EF)
∴ the events E and F are independent.
(ii) In a deck of 52 cards, 26 cards are black and 4 cards are kings.
26 1
∴ P(E) = P (the card drawn is a black) = \(\frac{26}{52}\) = \(\frac{1}{2}\)
Also P(F) = P (the card drawn in an ace) = \(\frac{4}{52}\) = \(\frac{1}{13}\)
In the pack of 52 cards, 2 cards are black as well as kings.
P(EF) = P( the card drawn is black king) = \(\frac{2}{52}\) = \(\frac{1}{26}\)
Now P(E)P(F) = \(\frac{1}{2}\) × \(\frac{1}{13}\) = \(\frac{1}{26}\) = P(EF)
∴ the given events E and F are independent.
(iii) In a deck of 52 cards, 4 cards are kings, 4 cards are queens, and 4 cards are jacks.
∴ P(E) = P(the card drawn is a king or a queen) = \(\frac{8}{52}\) = \(\frac{2}{13}\)
Also P(F) = P(the card drawn in a queen or a jack) = \(\frac{8}{52}\) = \(\frac{3}{13}\)
There are 4 cards which are king or queen and queen or jack.
P(EF) = P(the card drawn is king or a queen, or queen or a jack) = \(\frac{4}{52}\) = \(\frac{1}{13}\)
Now P(E)P(F) = \(\frac{2}{13}\) × \(\frac{2}{13}\) = \(\frac{4}{1}\) ≠ \(\frac{1}{13}\)
P(E)P(F) ≠ P(EF)
∴ the given events E and F are not independent.
Question 2.
In a hostel, 60% of the students read Hindi newspaper, 40% read English newspaper and 20% read both Hindi and English newspapers. A student is selected at random.
(a) Find the probability that she reads neither Hindi nor English newspapers.
(b) If she reads Hindi newspaper, find the probability that she reads English newspaper.
(c) If she reads English newspaper, find the probability that she reads Hindi newspaper.
Solution:
Let H denote the students who read Hindi newspaper and
E denote the students who read English newspaper.
Given that, P (H ) = 60% = \(\frac{60}{100}\) = \(\frac{3}{5}\)
P(E) = 40% = \(\frac{40}{100}\) = \(\frac{2}{5}\)
P(H ∩ E) = 20% = \(\frac{20}{100}\) = \(\frac{1}{5}\)
(i) Probability that a student reads neither Hindi nor English newspaper
P(H’ ∪ E’) = 1 – P(H ∪ E)
= 1 – [P(H) + P(E) – P(H ∩ E)]
= 1 – (\(\frac{3}{5}\) + \(\frac{2}{5}\) – \(\frac{1}{5}\))
= 1 – \(\frac{4}{5}\) = \(\frac{1}{5}\)
(ii) Probability that a randomly chosen student reads English newspaper, if she reads Hindi newspaper, is given by P(E|H)
P(E|H) = \(\frac{P(E \cap H)}{P(H)}\)
= \(\frac{\frac{1}{5}}{\frac{3}{5}}\) = \(\frac{1}{3}\)
(iii) Probability that a randomly chosen student reads Hindi newspaper, if she reads English newspaper, is given by P(H|E).
P(H|E) = \(\frac{P(H \cap E)}{P(E)}\)
= \(\frac{\frac{1}{5}}{\frac{2}{5}}\) = \(\frac{1}{2}\)