12th NCERT Probability Exercise 13.1 Questions 17
Hi

Question (1)

Given that E and F are events such that P(E) = 0.6, P(F) = 0.3 and P(E ∩ F) = 0.2, find P(E|F) and P(F|E).

Solution

$P(E/F) = \frac{{P(E \cap F)}}{{P(F)}}$ $= \frac{{0.2}}{{0.3}}$ $= \frac{2}{3}$ $P(F/E) = \frac{{P(E \cap F)}}{{P(E)}}$ $= \frac{{0.2}}{{0.6}}$ $= \frac{1}{3}$

Question (2)

Compute P(A|B), if P(B) = 0.5 and P(A ∩ B) = 0.32

Solution

$P\left( {A/B} \right) = \frac{{P\left( {A \cap B} \right)}}{{P\left( B \right)}}$ $= \frac{{0.32}}{{0.5}}$ $= 0.64$

Question (3)

If P(A) = 0.8, P(B) = 0.5 and P(B|A) = 0.4, find (i) P(A ∩ B) (ii) P(A|B) (iii) P(A ∪ B)

Solution

(i) $P\left( {B/A} \right) = \frac{{P\left( {A \cap B} \right)}}{{P\left( A \right)}}$ $0.4 = \frac{{P\left( {A \cap B} \right)}}{{0.8}}$ $P\left( {A \cap B} \right) = 0.4 \times 0.8$ $= 0.32$ (ii) $P\left( {A/B} \right) = \frac{{P\left( {A \cap B} \right)}}{{P\left( B \right)}}$ $= \frac{{0.32}}{{0.5}}$ $= 0.64$ (iii) $P\left( {A \cup B} \right) = P\left( A \right) + P\left( B \right) - P\left( {A \cap B} \right)$ $= 0.8 + 0.5 - 0.32$ $= 0.98$

Question (4)

Evaluate P(A ∪ B), if 2P(A) = P(B) = 5/13, and P(A|B) = 2/5.

Solution

$2P\left( A \right) = P\left( B \right) = \frac{5}{{13}}$ $\Rightarrow P\left( A \right) = \frac{5}{{26}}$ $P\left( {A/B} \right) = \frac{{P\left( {A \cap B} \right)}}{{P\left( B \right)}}$ $\frac{2}{5} = \frac{{P\left( {A \cap B} \right)}}{{5/13}}$ $P\left( {A \cap B} \right) = \frac{2}{5} \times \frac{5}{{13}}$ $= \frac{2}{{13}}$ $P\left( {A \cup B} \right) = P\left( A \right) + P\left( B \right) - P\left( {A \cap B} \right)$ $= \frac{5}{{26}} + \frac{5}{{13}} - \frac{2}{{13}}$ $= \frac{{5 + 10 - 4}}{{26}}$ $= \frac{{11}}{{26}}$

Question (5)

If P(A) = 6/11, P(B) = 5/11 and P(A ∪ B) = 7/11 , find
(𝑖) 𝑃(A ∩ 𝐵) (ii) 𝑃(𝐴|𝐵) (iii) 𝑃(𝐵|𝐴)

Solution

$P\left( {A \cup B} \right) = P\left( A \right) + P\left( B \right) - P\left( {A \cap B} \right)$ $\Rightarrow \frac{7}{{11}} = \frac{6}{{11}} + \frac{5}{{11}} - P\left( {A \cap B} \right)$ $P\left( {A \cap B} \right) = \frac{6}{{11}} + \frac{5}{{11}} - \frac{7}{{11}}$ $P\left( {A \cap B} \right) = \frac{4}{{11}}$ (ii) $P\left( {A/B} \right) = \frac{{P\left( {A \cap B} \right)}}{{P\left( B \right)}}$ $= \frac{{4/11}}{{5/11}}$ $= \frac{4}{5}$ (iii)$P\left( {B/A} \right) = \frac{{P\left( {A \cap B} \right)}}{{P\left( A \right)}}$ $= \frac{{4/11}}{{6/11}}$ $= \frac{4}{6} = \frac{2}{3}$ Determine P (E /F) for Q, 6 to 9

Question (6)

A coin is tossed three times , where
(i) E : head on third coin , F : heads on first two tosses
(ii) E : at least two heads, F : at most two heads
(iii) E : at the most two tails , F : at least one tail

Solution

A coin is tossed three times , So sample space will contain 8 elements.
S = { HHH, HHT, HTH, HTT, THH, THT, TTH, TTT}
So n = 8 .
(i) E : head on third coin.
∴ E = { HHH, HTH, THH, TTH }
m = 4. P( E) = 4/8 = 1/2.
F : Heads on first two tosses.
∴ F = { HHH, HHT } , m = 2
∴ P( F) = 2/8 = 1/4.
E ∩ F = { HHH}
∴ P (E ∩ F) = 1/8.
$P\left( {E/F} \right) = \frac{{P\left( {E \cap F} \right)}}{{P\left( F \right)}}$ $= \frac{{1/8}}{{2/8}}$ $= \frac{1}{2}$ (ii) E : at least two heads
E = { HHH,HHT, HTH, THH } , m = 4
P(E) = 4/8 = 1/2
F : at most two heads
F = { HHT, HTH, HTT, THH, THT, TTH, TTT } , m = 7
P(F) = 7/8.
E ∩ F = { HHT, HTH, THH } , m = 3
∴ P(E ∩ F) = 3/8.
$P\left( {E/F} \right) = \frac{{P\left( {E \cap F} \right)}}{{P\left( F \right)}}$ $= \frac{{3/8}}{{7/8}}$ $= \frac{3}{7}$ (iii) E : at most two tails
E = { HHH, HHT, HTH, THH, TTH, THT, HTT } , m = 7
P(E) = 7/8
F : at least one tail
F = { HHT, HTH, HTT, THH, THT, TTH, TTT } , m = 7
P(F) = 7/8.
E ∩ F = {HHT, HTH, THH, TTH, THT, HTT } , m = 6
∴ P(E ∩ F) = 6/8.
$P\left( {E/F} \right) = \frac{{P\left( {E \cap F} \right)}}{{P\left( F \right)}}$ $= \frac{{6/8}}{{7/8}}$ $= \frac{6}{7}$

Question (7)

Two coins are tossed once, where,
(i) E: tail appears on one coin, F: one coin shows head (ii) E: not tail appears, F: no head appears

Solution

Two coins are tossed once, So sample space will contain 4 elements.
S = { HH, HT, TH, TT}
(i) E: tail appears on one coin,
E = { HT, TH}, m = 2
P(E) = 2/4 = 1/2.
F: one coin shows head F = { HT , TH}, m = 2
P(F) = 2/4 = 1/2
E ∩ F = {HT, TH } m = 2
P(E ∩ F) = 2/4 = 1/2
$P\left( {E/F} \right) = \frac{{P\left( {E \cap F} \right)}}{{P\left( F \right)}}$ $= \frac{{1/2}}{{1/2}}$ $= 1$ (ii) E: not tail appears,
E = { HH}, m = 1
P(E) = 1/4 .
F: no head appears F = { TT}, m = 1
P(F) = 1/4
E ∩ F = { } m = 0
P(E ∩ F) = 0/4 = 0
$P\left( {E/F} \right) = \frac{{P\left( {E \cap F} \right)}}{{P\left( F \right)}}$ $= \frac{{0}}{{1/4}}$ $= 0$

Question (8)

. A die is thrown three times,
E: 4 appears on the third toss,
F: 6 and 5 appears respectively on first two tosses

Solution

A die is thrown three times. n = 63 = 216
E: 4 appears on the third toss, Any number can appear on first two die. ∴ m = 36,
P(E) = 36/216
F: 6 and 5 appears respectively on first two tosses.
First two dies have 6 and 5 the third die will have any number.
∴ m = 6
P(F) = 6/216
E ∩F = {(6, 5, 4)}
∴ P(E ∩ F) = 1/216
$P\left( {E/F} \right) = \frac{{P\left( {E \cap F} \right)}}{{P\left( F \right)}}$ $= \frac{{1/216}}{{6/216}}$ $= \frac{1}{6}$

Question (9)

Mother, father and son line up at random for a family picture
E: son on one end,
F: father in middle

Solution

As three persons line up in a row for family photo.
The number of arrangement possible = 3! = 6.
E: son on one end
For son there are two position and remaning two position are filled by parents in 2! ways.
so m = 2 × 2! = 4
E = { SMF, SFM, FMS, MFS} P(E) = 4/6
F: father in middle
For father there is only one position, and other two positions are filled by two persons in 2! ways.
∴ m = 1 ×2! = 2
F = { SFM, MFS} , P(F) = 2/6
E ∩F = { SFM, MFS} , m = 2
P(E ∩F) = 2/6
$P\left( {E/F} \right) = \frac{{P\left( {E \cap F} \right)}}{{P\left( F \right)}}$ $= \frac{{2/6}}{{2/6}}$ $= 1$

Question (10)

A black and a red dice are rolled. (a) Find the conditional probability of obtaining a sum greater than 9, given that the black die resulted in a 5. (b) Find the conditional probability of obtaining the sum 8, given that the red die resulted in a number less than 4.

Solution

A black and red die is rolled . Sample space will contain 36 elements. n = 36
(i) A : sum is greater than 9.
A = { (4,6),(5,5),(6,4),(5,6),(6,5),(6,6)} P(A) = 6/36.
B : A black die resulted in 5. i.e. first die shows 5.
B = { (5,1),(5,2),(5,3),(5,4),(5,5),(5,6)}
P(B) = 6/36
A ∩ B = { (5,5),((5,6)}, m = 2
P(A ∩ B) = 2/36
$P\left( {A/B} \right) = \frac{{P\left( {A \cap B} \right)}}{{P\left( B \right)}}$ $= \frac{{2/36}}{{6/36}}$ $= \frac{1}{3}$ (ii) A sum is 8
A = {(2,6),(3,5),(4,4),(5,3)(6,2)}
P(A) = 5/36.
B : Red die resuled in a number less than 4, That is second die shows the number less than 4.
m = 18, ∴P(B) = 18/36.
A∩B = { (5,3),(6,2)} , m = 2
P(A ∩ B) = 2/36
$P\left( {A/B} \right) = \frac{{P\left( {A \cap B} \right)}}{{P\left( B \right)}}$ $= \frac{{2/36}}{{18/36}}$ $= \frac{1}{9}$

Question (11)

A fair die is rolled. Consider events E = {1,3, 5}, F = {2, 3} and G = {2, 3,4, 5} Find (i) P(E|F) and P(F|E) (ii) P(E|G) and P(G|E) (iii) P((E ∪ F)|G) and P((E ∩ F)|G)

Solution

A fair die is rolled , so n = 6.
E = {1,3, 5}, P(E) = 3/6
F = {2, 3}, P(F) = 2/6
and G = {2, 3,4, 5}.P(G) = 4/6
(i) E ∩F = {3} , P(E ∩F) = 1/6
$P\left( {E/F} \right) = \frac{{P\left( {E \cap F} \right)}}{{P\left( F \right)}}$ $= \frac{{1/6}}{{2/6}}$ $= \frac{1}{2}$ $P\left( {F/E} \right) = \frac{{P\left( {E \cap F} \right)}}{{P\left( E \right)}}$ $= \frac{{1/6}}{{3/6}}$ $= \frac{1}{3}$ (ii) E ∩G = { 3, 5} , P(E ∩G) = 2/6
$P\left( {E/G} \right) = \frac{{P\left( {E \cap G} \right)}}{{P\left( G \right)}}$ $= \frac{{2/6}}{{4/6}}$ $= \frac{1}{2}$ $P\left( {G/E} \right) = \frac{{P\left( {E \cap G} \right)}}{{P\left( E \right)}}$ $= \frac{{2/6}}{{3/6}}$ $= \frac{2}{3}$ (iii)( E ∪F ) = { 1, 2, 3, 5} , P(E ∪ F) = 4/6.
( (E ∪F) ∩ G) = { 2, 3, 5} , P(( E ∪F) ∩ G) = 3/6
$P\left( {E \cup F/G} \right) = \frac{{P\left[ {\left( {E \cup F} \right) \cap G} \right]}}{{P\left( G \right)}}$ $= \frac{{3/6}}{{4/6}}$ $= \frac{{3}}{{4}}$ (E ∩ F)= { 3} , P(E ∩ G) = 1/6
(E ∩ F)∩ G = {3} , P((E ∩ F)∩ G) = 1/6
$P\left( {E \cap F/G} \right) = \frac{{P\left[ {\left( {E \cap F} \right) \cap G} \right]}}{{P\left( G \right)}}$ $= \frac{{1/6}}{{4/6}}$ $= \frac{1}{4}$

Question (12)

Assume that each born child is equally likely to be a boy or a girl. If a family has two children, what is the conditional probability that both are girls given that (i) the youngest is a girl, (ii) at least one is a girl?

Solution

A family has two children. { BB, BG, GB, GG}. n = 4.
Let A : both are girls.
A = {GG}, P(A) = 1/4.
(i) B : the youngest is a girl.
B = {BG, GG} , P(B) = 2/4.
A ∩B = {GG}, P(A ∩B) = 1/4
$P\left( {A/B} \right) = \frac{{P\left[ {A \cap B} \right]}}{{P\left( B \right)}}$ $= \frac{{1/4}}{{2/4}}$ $= \frac{1}{2}$ C : At least one is girl
C = { BG,GB,GG} , P(C) = 3/4.
A ∩C = {GG}, P(A ∩C) = 1/4
$P\left( {A/C} \right) = \frac{{P\left[ {A \cap C} \right]}}{{P\left( C \right)}}$ $= \frac{{1/4}}{{3/4}}$ $= \frac{1}{3}$

Question (13)

An instructor has a question bank consisting of 300 easy True/False Questions, 200 difficult True/False Questions, 500 easy multiple choice questions and 400 difficult multiple choice questions. If a question is selected at random from the question bank, what is the probability that it will be an easy question given that it is a multiple choice question?

Solution

From the given information we will prepare the table as follows.
 MCQ True/ false Total Easy 500 300 800 Difficult 400 200 600 total 900 500 1400

Let A : the selected question is easy.
P(A) = 800/1400.
Let B : The selected question is MCQ.
P(B) = 900/ 1400
P( A ∩ B ) = 500/1400
$P\left( {A/B} \right) = \frac{{P\left[ {A \cap B} \right]}}{{P\left( B \right)}}$ $= \frac{{500/1400}}{{900/1400}}$ $= \frac{5}{9}$

Question (14)

Given that the two numbers appearing on throwing the two dice are different. Find the probability of the event ‘the sum of numbers on the dice is 4’.

Solution

The two dices are thrown , n = 36.
Let A : The numbers are different.
P(A) = 30/36.
Let B : The sum of numbers on dice is 4.
B = {(1,3),(2,2),(3,1)}
P(B) = 3 / 36.
A ∩B = { (1,3),(3,1)}.
P(A ∩ B) = 2/36.
$P\left( {B/A} \right) = \frac{{P\left[ {A \cap B} \right]}}{{P\left( A \right)}}$ $= \frac{{2/36}}{{30/36}}$ $= \frac{1}{{15}}$

Question (15)

Consider the experiment of throwing a die, if a multiple of 3 comes up, throw the die again and if any other number comes, toss a coin. Find the conditional probability of the event ‘the coin shows a tail’, given that ‘at least one die shows a 3’.

Solution

Consider the experiment of throwing a die, if a multiple of 3 comes up, throw the die again and if any other number comes, toss a coin. Sample space will contain 20 elements. n = 20.
Let A be the event coin shows tail
m = 4, P(A) = 4/20.
Let B be the event At least one die shows 3.
m = 7, P(B) = 7/20.
A ∩ B is an impossible event .
∴ P(A ∩B) = 0.
$P\left( {A/B} \right) = \frac{{P\left[ {A \cap B} \right]}}{{P\left( B \right)}}$ $= \frac{{0}}{{7/20}}$ $= 0$

Question (16)

. If P(A) =1/2, P(B) = 0, then P(A|B) is
(A) 0 (B) 1/2 (C) not defined (D) 1

Solution

Since P(B) = 0,
$P\left( {A/B} \right) = \frac{{P\left[ {A \cap B} \right]}}{{P\left( B \right)}}$ $= \frac{{P\left[ {A \cap B} \right]}}{0}$ So it is not defined.
So C is the correct answer.

Question (17)

If A and B are events such that P(A|B) = P(B|A), then
(A) A ⊂ B but A ≠ B (B) A = B (C) A ∩ B = Φ (D) P(A) = P(B)

Solution

$P\left( {A/B} \right) = \frac{{P\left[ {A \cap B} \right]}}{{P\left( B \right)}}$ $P\left( {B/A} \right) = \frac{{P\left[ {A \cap B} \right]}}{{P\left( A \right)}}$ P(A/B) = P(B/A)
⇒ P(A) = P(B)
So D is the correct answer.