12th NCERT Probability Miscallneous Exercise 13 Questions 19
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Question (1)

A and B are two events such that P(A) ≠ 0. Find P(B|A), if (i) A is a subset of B (ii) A ∩ B = Φ

Solution

(i) A is the subset of B,
∴ A ∩ B = A, P( A ∩ B) = P(A)
$P\left( {B/A} \right) = \frac{{P\left( {A \cap B} \right)}}{{P\left( A \right)}}$ $= \frac{{P\left( A \right)}}{{{\rm{ P}}\left( {\rm{A}} \right)}}$ $= 1$ (ii) A ∩ B = Φ , P( A ∩ B) = 0
$P\left( {B/A} \right) = \frac{{P\left( {A \cap B} \right)}}{{P\left( A \right)}}$ $= \frac{0}{{{\rm{ P}}\left( {\rm{A}} \right)}}$ $= 0$

Question (2)

A couple has two children, (i) Find the probability that both children are males, if it is known that at least one of the children is male. (ii) Find the probability that both children are females, if it is known that the elder child is a female.

Solution

A cople has 2 children, Sample space S = { BB, BG, GB, GG}
(i) Let A be the event both are male , A = {BB}, P(A) = 1/4.
Let B be the event that atleast one of the children is male
B = { BB, BG, GB} , P(B) = 3/4.
A ∩ B = { BB) , P( A ∩ B) = 1/4.
$P\left( {A/B} \right) = \frac{{P\left( {A \cap B} \right)}}{{P\left( A \right)}}$ $= \frac{{1/4}}{{{\rm{ 3/4}}}}$ $= \frac{1}{3}$ (ii) Let A be the event both are female , A = {GG}, P(A) = 1/4.
Let B be the event that elder child is female
B = { GB, GG} , P(B) = 2/4.
A ∩ B = { GG) , P( A ∩ B) = 1/4.
$P\left( {A/B} \right) = \frac{{P\left( {A \cap B} \right)}}{{P\left( A \right)}}$ $= \frac{{1/4}}{{{\rm{ 2/4}}}}$ $= \frac{1}{2}$

Question (3)

Suppose that 5% of men and 0.25% of women have grey hair. A grey haired person is selected at random. What is the probability of this person being male? Assume that there are equal number of males and females.

Solution

Let A is the event that selected person have gray hair.
Let E 1 is the event that selected person is male.
Let E 2 is the event that selected person is female.
$P\left( {{E_1}} \right) = P\left( {{E_2}} \right) = \frac{1}{2}$ $P\left( {A/{E_1}} \right) = 0.05,P\left( {A/{E_2}} \right) = 0.0025$ $P\left( {{E_1}/A} \right) = \frac{{P\left( {{E_1} \cap A} \right)}}{{P\left( A \right)}}$ $= \frac{{P\left( {{E_1}} \right)P\left( {A/{E_1}} \right)}}{{P\left( {{E_1}} \right)P\left( {A/{E_1}} \right) + P\left( {{E_2}} \right)P\left( {A/{E_2}} \right)}}$ $= \frac{{0.5 \times 0.05}}{{0.5 \times 0.05 + 0.5 \times 0.0025}}$ $= \frac{{0.025}}{{0.025 + 0.00125}}$ $= \frac{{0.025}}{{0.02625}}$ $= \frac{{2500}}{{2625}}$ $= \frac{{20}}{{21}}$

Question (4)

Suppose that 90% of people are right-handed. What is the probability that at most 6 of a random sample of 10 people are right-handed?

Solution

p = probability that person is right handed.
p = 9/10, q = 1 - p = 1/10, n = 10.
$P\left( {X = x} \right) = n{C_x}{p^x}{q^{n - x}}$ P( at most 6) = P( X ≤ 6)
P(X=0)+P(X=1)+P(X=2)+P(x=3)+P(X=4)+P(X=5)+P(X=6)
$= \sum\limits_0^6 {10{C_x}{{\left( {0.9} \right)}^x}{{\left( {0.1} \right)}^{10 - x}}}$

Question (5)

. An urn contains 25 balls of which 10 balls bear a mark ‘X’ and the remaining 15 bear a mark ‘Y’. A ball is drawn at random from the urn, its mark is noted down and it is replaced. If 6 balls are drawn in this way, find the probability that (i) all will bear ‘X’ mark. (ii) not more than 2 will bear ‘Y’ mark. (iii) at least one ball will bear ‘Y’ mark (iv) the number of balls with ‘X’ mark and ‘Y’ mark will be equal.

Solution

p = probability og getting a ball 'X'
p = 2/5, q = 1 - p = 3/5, n = 6
$P\left( {X = x} \right) = n{C_x}{p^x}{q^{n - x}}$ (i) P( X = 6) $= 6{C_6}{\left( {\frac{2}{5}} \right)^6}{\left( {\frac{3}{5}} \right)^0}$ $= 1 \times {\left( {\frac{2}{5}} \right)^6} \times 1$ $= {\left( {\frac{2}{5}} \right)^6}$ (ii) not more than 2 will bear 'Y' mark.
P( P(X ≥4)
= ( X =4) + P( X = 5 ) + P( X = 6) $= 6{C_4}{\left( {\frac{2}{5}} \right)^4}{\left( {\frac{3}{5}} \right)^2} + 6{C_5}{\left( {\frac{2}{5}} \right)^5}{\left( {\frac{3}{5}} \right)^1} + 6{C_6}{\left( {\frac{2}{5}} \right)^6}{\left( {\frac{3}{5}} \right)^0}$ $= {\left( {\frac{2}{5}} \right)^4}\left[ {15 \times \frac{9}{{25}} + 6 \times \frac{2}{5} \times \frac{3}{5} + 1 \times \frac{4}{{25}} \times 1} \right]$ $= {\left( {\frac{2}{5}} \right)^4}\left[ {\frac{{135 + 36 + 4}}{{25}}} \right]$ $= {\left( {\frac{2}{5}} \right)^4}\left[ {\frac{{175}}{{25}}} \right]$ $= 7{\left( {\frac{2}{5}} \right)^4}$ (iii)at least one ball bears 'Y' mark
= P(X ≤ 5 )
= P(X=0)+P(X=1)+P(X=2)+P(X=3)+P(X=4)+P(X=5)
= 1 - P( X = 6)
$= 1 - 6{C_6}{\left( {\frac{2}{5}} \right)^6}{\left( {\frac{3}{5}} \right)^0}$ $1 - 1 \times {\left( {\frac{2}{5}} \right)^6} \times 1$ $= 1 - {\left( {\frac{2}{5}} \right)^6}$ (iv) P(X = 3) $= 6{C_3}{\left( {\frac{2}{5}} \right)^3}{\left( {\frac{3}{5}} \right)^3}$ $= 20 \times \frac{8}{{125}} \times \frac{{27}}{{125}}$ $= \frac{{864}}{{3125}}$

Question (6)

In a hurdle race, a player has to cross 10 hurdles. The probability that he will clear each hurdle is 5/6. What is the probability that he will knock down fewer than 2 hurdles?

Solution

p = probability of clearing the hurdle.
p = 5/6, q = 1 - p = 1/6, n = 10
$P\left( {X = x} \right) = n{C_x}{p^x}{q^{n - x}}$ the probability that he will knock down fewer than 2 hurdles?
= P( X >8)
= P( X = 9) + P( X = 10)
$= 10{C_9}{\left( {\frac{5}{6}} \right)^9}{\left( {\frac{1}{6}} \right)^1} + 10{C_{10}}{\left( {\frac{5}{6}} \right)^{10}}{\left( {\frac{1}{6}} \right)^0}$ $= {\left( {\frac{5}{6}} \right)^9}\left[ {10 \times \frac{1}{6} + 1 \times \frac{5}{6}} \right]$ $= {\left( {\frac{5}{6}} \right)^9}\left[ {\frac{{10 + 5}}{{26}}} \right]$ $= {\left( {\frac{5}{6}} \right)^9}\left[ {\frac{{15}}{6}} \right]$ $= \frac{5}{2}{\left( {\frac{5}{6}} \right)^4}$

Question (7)

A die is thrown again and again until three sixes are obtained. Find the probability of obtaining the third six in the sixth throw of the die.

Solution

A die is thrown again and again until three sixes are obtained. The third six we get in the sixth draw.It indicate we get two sixes in first five draws.
Let p = probability getting six when die is thrown .
$P\left( {X = x} \right) = n{C_x}{p^x}{q^{n - x}}$ p = 1/6, q = 1 - p = 5/6, n = 5.
P( third six in sixth draw)
= P( 2 sixes in 5 draws ) × P( getting 6 in the sixth draw)
$= 5{C_2}{\left( {\frac{1}{6}} \right)^2}{\left( {\frac{5}{6}} \right)^3} \times \frac{1}{6}$ $= 10 \times \frac{1}{{36}} \times \frac{{125}}{{216}} \times \frac{1}{6}$ $= \frac{{625}}{{23328}}$

Question (8)

If a leap year is selected at random, what is the chance that it will contain 53 tuesdays?

Solution

In a leap year there are 366 days.
That is it have 52 weeks and 2 days.
To get 53 tuesdays, this extra day should have tuesday.
The Sample space for these two days is
S = { MT, TW, Wt, tF, FS, Ss, sM} where T is tuesday, t = thursday, S = saturday, s = sunday.
n = 7.
A ; One day among these two days is tuesday, A = {MT, TW}, m = 2
P( A) = m/n.
= 2/7.

Question (9)

An experiment succeeds twice as often as it fails. Find the probability that in the next six trials, there will be at least 4 successes.

Solution

An experiment succeeds twice as often as it fails.
p = 2q
p + q = 1
2q + q = 1
3q = 1 ⇒ q = 1/3
p = 2/3, n = 6
P( atleast 4 success)
P( X = 4) + P( x = 5 ) + P( X = 6)
$P\left( {X = x} \right) = n{C_x}{p^x}{q^{n - x}}$ $= 6{C_4}{\left( {\frac{2}{3}} \right)^4}{\left( {\frac{1}{3}} \right)^2} + 6{C_5}{\left( {\frac{2}{3}} \right)^5}{\left( {\frac{1}{3}} \right)^1} + 6{C_6}{\left( {\frac{2}{3}} \right)^6}{\left( {\frac{1}{3}} \right)^0}$ $= {\left( {\frac{2}{3}} \right)^4}\left[ {15 \times \frac{1}{9} + 6 \times \frac{2}{3} \times \frac{1}{3} + 1 \times \frac{4}{9}} \right]$ $= {\left( {\frac{2}{3}} \right)^4}\left[ {\frac{{15 + 12 + 4}}{9}} \right]$ $= {\left( {\frac{2}{3}} \right)^4}\left[ {\frac{{31}}{9}} \right]$

Question (10)

How many times must a man toss a fair coin so that the probability of having at least one head is more than 90%?

Solution

p = getting head when coin is tossed.
p = 1/2, q = 1 - p = 1/2, n = ?
P ( X ≥ 1) >0.9
$P\left( {X = x} \right) = n{C_x}{p^x}{q^{n - x}}$ P ( X ≥ 1) >0.9
1 - P( X = 0 ) > 0.90
0 .10 > P( X = 0 )
P( X = 0) < 0.1
$n{C_0}{\left( {\frac{1}{2}} \right)^0}{\left( {\frac{1}{2}} \right)^n} < 0.1$ $1 \times 1 \times {\left( {\frac{1}{2}} \right)^n} < 0.1$ ${\left( {\frac{1}{2}} \right)^n} < 0.1$ ${2^n} > \frac{1}{{0.1}} > 10$ $\Rightarrow n \ge 4$

Question (11)

In a game, a man wins a rupee for a six and loses a rupee for any other number when a fair die is thrown. The man decided to throw a die thrice but to quit as and when he gets a six. Find the expected value of the amount he wins/loses.

Solution

A man wins Rs. 1 if he get number six . Probability of getting six is 1/6, probability of not getting six is 5/6.
If person may get six in second draw , then its probability is 5/36, he will not get any money.
X ; the amount the person will earn.
We will prepare the table as follows.
 X P(X) XP(X) 1 1/6 1/6 0 5/36 0 -1 25/216 -25/216 -3 125/216 -375/216 Total 1 -364/216

E(X) = ∑XP(X)
= -364/ 216
= - 91/54 . He will lose Rs. 91/54.

Question (12)

Suppose we have four boxes. A, B, C and D containing coloured marbles as given below:
 Box Red marbles White marbles Black marbles A 1 6 3 B 6 2 2 C 8 1 1 D 0 6 4

One of the boxes has been selected at random and a single marble is drawn from it. If the marble is red, what is the probability that it was drawn from box A?, box B?, box C?

Solution

Let A be the event that selected marble is red.
E1 be the event that marble is selected from box A.
E2 be the event that marble is selected from box B.
E3 be the event that marble is selected from box C.
E4 be the event that marble is selected from box D.
$P\left( {{E_1}} \right) = \frac{1}{4},P\left( {{E_2}} \right) = \frac{1}{4},P\left( {{E_3}} \right) = \frac{1}{4},P\left( {{E_4}} \right) = \frac{1}{4}$ $P\left( {A/{E_1}} \right) = \frac{1}{{10}},P\left( {A/{E_2}} \right) = \frac{6}{{10}},P\left( {A/{E_2}} \right) = \frac{8}{{10}},P\left( {A/{E_2}} \right) = \frac{0}{{10}}$ (i) Marble is from box A.
$P\left( {{E_1}/A} \right) = \frac{{P\left( {{E_1} \cap A} \right)}}{{P\left( A \right)}}$ $= \frac{{P\left( {{E_1}} \right)P\left( {A/{E_1}} \right)}}{{P\left( {{E_1}} \right)P\left( {A/{E_1}} \right) + P\left( {{E_2}} \right)P\left( {A/{E_2}} \right) + P\left( {{E_3}} \right)P\left( {A/{E_3}} \right) + P\left( {{E_4}} \right)P\left( {A/{E_4}} \right)}}$ $= \frac{{\frac{1}{4} \times \frac{1}{{10}}}}{{\frac{1}{4} \times \frac{1}{{10}} + \frac{1}{4} \times \frac{6}{{10}} + \frac{1}{4} \times \frac{8}{{10}} + \frac{1}{4} \times \frac{0}{{10}}}}$ $= \frac{{\frac{1}{{40}}}}{{\frac{{1 + 6 + 8 + 0}}{{40}}}}$ $= \frac{1}{{15}}$ (ii) Marble is from box B.
$P\left( {{E_2}/A} \right) = \frac{{P\left( {{E_2} \cap A} \right)}}{{P\left( A \right)}}$ $= \frac{{P\left( {{E_2}} \right)P\left( {A/{E_2}} \right)}}{{P\left( {{E_1}} \right)P\left( {A/{E_1}} \right) + P\left( {{E_2}} \right)P\left( {A/{E_2}} \right) + P\left( {{E_3}} \right)P\left( {A/{E_3}} \right) + P\left( {{E_4}} \right)P\left( {A/{E_4}} \right)}}$ $= \frac{{\frac{1}{4} \times \frac{6}{{10}}}}{{\frac{1}{4} \times \frac{1}{{10}} + \frac{1}{4} \times \frac{6}{{10}} + \frac{1}{4} \times \frac{8}{{10}} + \frac{1}{4} \times \frac{0}{{10}}}}$ $= \frac{{\frac{6}{{40}}}}{{\frac{{1 + 6 + 8 + 0}}{{40}}}}$ $= \frac{6}{{15}}$ $= \frac{2}{{5}}$ (iii) Marble is from box C
$P\left( {{E_3}/A} \right) = \frac{{P\left( {{E_3} \cap A} \right)}}{{P\left( A \right)}}$ $= \frac{{P\left( {{E_3}} \right)P\left( {A/{E_3}} \right)}}{{P\left( {{E_1}} \right)P\left( {A/{E_1}} \right) + P\left( {{E_2}} \right)P\left( {A/{E_2}} \right) + P\left( {{E_3}} \right)P\left( {A/{E_3}} \right) + P\left( {{E_4}} \right)P\left( {A/{E_4}} \right)}}$ $= \frac{{\frac{1}{4} \times \frac{8}{{10}}}}{{\frac{1}{4} \times \frac{1}{{10}} + \frac{1}{4} \times \frac{6}{{10}} + \frac{1}{4} \times \frac{8}{{10}} + \frac{1}{4} \times \frac{0}{{10}}}}$ $= \frac{{\frac{8}{{40}}}}{{\frac{{1 + 6 + 8 + 0}}{{40}}}}$ $= \frac{8}{{15}}$

Question (13)

Assume that the chances of a patient having a heart attack are 40%. It is also assumed that a meditation and yoga course reduce the risk of heart attack by 30% and prescription of certain drug reduces its chances by 25%. At a time a patient can choose any one of the two options with equal probabilities. It is given that after going through one of the two options the patient selected at random suffers a heart attack. Find the probability that the patient followed a course of meditation and yoga?

Solution

Let A be the event the patient selected suffers from heart attack.
Let E1 be the event that person select Meditation and yoga .
Let E2 be the event that person select presciption. .
$P\left( {{E_1}} \right) = \frac{1}{2},P\left( {{E_2}} \right) = \frac{1}{2}$ Due to tretemant risk is reduce by certain percentage.
$P\left( {A/{E_1}} \right) = 0.4 - 0.3 \times 0.4 = 0.4 - 0.12 = 0.28$ $P\left( {A/{E_2}} \right) = 0.4 - 0.25 \times 0.4 = 0.4 - 0.1 = 0.3$ By Baye's theorem, $P\left( {{E_1}/A} \right) = \frac{{P\left( {{E_1} \cap A} \right)}}{{P\left( A \right)}}$ $= \frac{{P\left( {{E_1}} \right)P\left( {A/{E_1}} \right)}}{{P\left( {{E_1}} \right)P\left( {A/{E_1}} \right) + P\left( {{E_2}} \right)P\left( {A/{E_2}} \right)}}$ $= \frac{{\frac{1}{2} \times \frac{{28}}{{100}}}}{{\frac{1}{2} \times \frac{{28}}{{100}} + \frac{1}{2} \times \frac{{30}}{{100}}}}$ $= \frac{{\frac{{28}}{{200}}}}{{\frac{{28 + 30}}{{200}}}}$ $= \frac{{28}}{{58}}$ $= \frac{{14}}{{29}}$

Question (14)

If each element of a second order determinant is either zero or one, what is the probability that the value of the determinant is positive? (Assume that the individual entries of the determinant are chosen independently, each value being assumed with probability 1/2).

Solution

The determinant is of order 2 × 2, so it has 4 elements.
We are selecting the elements from the digit 0, and 1.
So possible way it can be done n = 2 4 = 16.
As the value of D is positive there are only possible D are
$\left| {\begin{array}{*{20}{c}}1&0\\0&1\end{array}} \right|,\left| {\begin{array}{*{20}{c}}1&1\\0&1\end{array}} \right|,\left| {\begin{array}{*{20}{c}}1&0\\1&1\end{array}} \right|$
so m = 3.
P(A) = m/n.
= 3/14.

Question (15)

An electronic assembly consists of two subsystems, say, A and B. From previous testing procedures, the following probabilities are assumed to be known: P(A fails) = 0.2 P(B fails alone) = 0.15 P(A and B fail) = 0.15 Evaluate the following probabilities (i) P(A fails|B has failed) (ii) P(A fails alone)

Solution

P( B fails) = P( B fails alone ) + P( A and B fails)
= 0.15 + 0.15
= 0.30
(i) P(A fails|B has failed) $= \frac{{P\left( {A\,and\,B\,fails} \right)}}{{P\left( {B\,fails} \right)}}$ $= \frac{{0.15}}{{0.30}}$ $= 0.5$ (ii) P(A fails alone)
= P(A fails) - P(A and B fail)
= 0.2 - 0.15
= 0.05

Question (16)

Bag I contains 3 red and 4 black balls and Bag II contains 4 red and 5 black balls. One ball is transferred from Bag I to Bag II and then a ball is drawn from Bag II. The ball so drawn is found to be red in colour. Find the probability that the transferred ball is black.

Solution

Let A be the event that ball drawn from bag II is red.
E1 be the event that red ball is transferred from bag I to bag II.
E2 be the event that black ball is transferred from bag I to bag II.
$P\left( {{E_1}} \right) = \frac{3}{7},P\left( {{E_2}} \right) = \frac{4}{7}$ $P\left( {A/{E_1}} \right) = \frac{5}{{10}},P\left( {A/{E_2}} \right) = \frac{4}{{10}}$ By Baye's theorem, $P\left( {{E_2}/A} \right) = \frac{{P\left( {{E_2} \cap A} \right)}}{{P\left( A \right)}}$ $= \frac{{P\left( {{E_2}} \right)P\left( {A/{E_2}} \right)}}{{P\left( {{E_1}} \right)P\left( {A/{E_1}} \right) + P\left( {{E_2}} \right)P\left( {A/{E_2}} \right)}}$ $= \frac{{\frac{4}{7} \times \frac{4}{{10}}}}{{\frac{3}{7} \times \frac{5}{{10}} + \frac{4}{7} \times \frac{4}{{10}}}}$ $= \frac{{\frac{{16}}{{70}}}}{{\frac{{15 + 16}}{{70}}}}$ $= \frac{{16}}{{31}}$

Question (17)

If A and B are two events such that P(A) ≠ 0 and P(B|A) = 1, then. (A) A ⊂ B (B) B ⊂ A (C) B = Φ (D) A = Φ

Solution

If A and B are two events such that P(A) ≠ 0 $P\left( {B/A} \right) = 1$ $\frac{{P\left( {A \cap B} \right)}}{{P\left( A \right)}} = 1$ $P\left( {A \cap B} \right) = P\left( A \right)$ $\Rightarrow A \subset B$ So A is the correct answer.

Question (18)

If P(A|B) > P(A), then which of the following is correct: (A) P(B|A) < P(B) (B) P(A ∩ B) < P(A). P(B) (C) P(B|A) > P(B) (D) P(B|A) = P(B)

Solution

If P(A|B) > P(A), $\frac{{P\left( {A \cap B} \right)}}{{P\left( B \right)}} > P\left( A \right)$ $P\left( {A \cap B} \right) > P\left( B \right)P\left( A \right)$ $P\left( {B/A} \right) = \frac{{P\left( {A \cap B} \right)}}{{P\left( A \right)}}$ $P\left( {B/A} \right) > \frac{{P\left( A \right)P\left( B \right)}}{{P\left( A \right)}}$ $P\left( {B/A} \right) > P\left( B \right)$ So C is the correct answer.

Question (19)

. If A and B are any two events such that P(A) + P(B) − P(A and B) = P(A), then (A) P(B|A) = 1 (B) P(A|B) = 1 (C) P(B|A) = 0 (D) P(A|B) = 0

Solution

If A and B are any two events such that P(A) + P(B) − P(A and B) = P(A)
⇒ P(B) = P( A and B) = P( A ∩B)
$P\left( {A/B} \right) = \frac{{P\left( {A \cap B} \right)}}{{P\left( B \right)}}$ $= \frac{{P\left( B \right)}}{{P\left( B \right)}}$ $= 1$ So B is the correct answer.