Probability — Page 5 | JEE Mathematics

Q41

A dice is thrown two times and the sum of the scores appearing on the die is observed to be a multiple of 4. Then the conditional probability that the score 4 has appeared atleast once is :

A

{1 \over 8}

B

{1 \over 9}

C

{1 \over 4}

D

{1 \over 3}

Correct Answer

Option B

Solution

Let A is the event for getting score a multiple of 4.

So, A = { (1, 3), (3, 1), (2, 2), (2, 6), (6, 2), (3, 5), (5, 3), (4, 4), (6, 6) } = 09 n(A) = 9 B : Score of 4 has appeared at least once.

B = {(4, 4)} So, Required probability =

{1 \over 9}

Q42

Box I contains 30 cards numbered 1 to 30 and Box II contains 20 cards numbered 31 to 50. A box is selected at random and a card is drawn from it. The number on the card is found to be a non-prime number. The probability that the card was drawn from Box I is :

A

{8 \over {17}}

B

{2 \over 3}

C

{2 \over 5}

D

{4 \over {17}}

Correct Answer

Option A

Solution

Let B1 be the event where Box-I is selected. And B2 be the event where Box-II is selected. P(B1) = P(B2) =

{1 \over 2}

Let E be the event where selected card is non prime.

For B1 : Prime numbers: {2, 3, 5, 7, 11, 13, 17, 19, 23, 29} For B2 : Prime numbers: {31, 37, 41, 43, 47} P(E) = P(B1) $\times$

P\left( {{E \over {{B_1}}}} \right)

+ P(B2) $\times$

P\left( {{E \over {{B_2}}}} \right)

=

{1 \over 2} \times {{20} \over {30}}

+

{1 \over 2} \times {{15} \over {20}}

Required probability :

P\left( {{{{B_1}} \over E}} \right)

=

{{P\left( {{B_2}} \right).P\left( {{E \over {{B_1}}}} \right)} \over {P\left( E \right)}}

=

{{{1 \over 2} \times {{20} \over {30}}} \over {{1 \over 2} \times {{20} \over {30}} + {1 \over 2}{{15} \over {20}}}}

=

{{{2 \over 3}} \over {{2 \over 3} + {3 \over 4}}}

=

{8 \over {17}}

Q43

If 10 different balls are to be placed in 4 distinct boxes at random, then the probability that two of these boxes contain exactly 2 and 3 balls is :

A

{{965} \over {{2^{11}}}}

B

{{965} \over {{2^{10}}}}

C

{{945} \over {{2^{11}}}}

D

{{945} \over {{2^{10}}}}

Correct Answer

Option D

Solution

Total ways of distribution = 410 = 220 Number of ways selecting two boxes out of four = 4C2 Then number of ways selecting 5 balls out of 10 = 10C5 Then no of ways of distributing 5 balls into two groups of 2 balls and 3 balls = 5C3.2!

Then number of ways to distributing remaining balls into two boxes = 25 Number of ways placing exactly 2 and 3 balls in two of these boxes = 4C2 $\times$ 10C5 $\times$ 5C3.2!

$\times$ 25 =

{{{{6.252.10.2.2}^5}} \over {{2^{20}}}}

=

{{945} \over {{2^{10}}}}

Q44

In a box, there are 20 cards, out of which 10 are lebelled as A and the remaining 10 are labelled as B. Cards are drawn at random, one after the other and with replacement, till a second A-card is obtained. The probability that the second A-card appears before the third B-card is :

A

{{13} \over {16}}

B

{{11} \over {16}}

C

{{15} \over {16}}

D

{{9} \over {16}}

Correct Answer

Option B

Solution

Possibilities that the second A card appears before the third B card are =AA + ABA + BAA + ABBA + BBAA + BABA =

{\left( {{1 \over 2}} \right)^2}

+

{\left( {{1 \over 2}} \right)^3}

+

{\left( {{1 \over 2}} \right)^3}

+

{\left( {{1 \over 2}} \right)^4}

+

{\left( {{1 \over 2}} \right)^4}

+

{\left( {{1 \over 2}} \right)^4}

=

{1 \over 4}

+

{1 \over 8}

+

{1 \over 8}

+

{1 \over {16}}

+

{1 \over {16}}

+

{1 \over {16}}

=

{{11} \over {16}}

Q45

Let A and B be two events such that the probability that exactly one of them occurs is

{2 \over 5}

and the probability that A or B occurs is

{1 \over 2}

, then the probability of both of them occur together is :

A 0.20

B 0.02

C 0.01

D 0.10

Correct Answer

Option D

Solution

Probability that exactly one of them occurs P(A) + P(B) – 2P (A $\cap$ B) =

{2 \over 5}

.....(1) Probability that A or B occurs is P(A) + P(B) – P(A $\cap$ B) =

{1 \over 2}

......(2) Doing (2) - (1) P(A $\cap$ B) =

{1 \over 2} - {2 \over 5}

= 0.10

Q46

Let A and B be two independent events such that P(A) =

{1 \over 3}

and P(B) =

{1 \over 6}

. Then, which of the following is TRUE?

A

P\left( {{A \over {A \cup B}}} \right) = {1 \over 4}

B

P\left( {{A \over B}} \right) = {2 \over 3}

C

P\left( {{{A'} \over {B'}}} \right) = {1 \over 3}

D

P\left( {{A \over {B'}}} \right) = {1 \over 3}

Correct Answer

Option D

Solution

Given P(A) =

{1 \over 3}

and P(B) =

{1 \over 6}

A and B are independent So P(A $\cap$ B) =

{1 \over 3} \times {1 \over 6}

=

{1 \over {18}}

P(A $\cup$ B) = P(A) + P(B) – P(A $\cap$ B) =

{1 \over 3}

+

{1 \over 6}

-

{1 \over {18}}

=

{4 \over 9}

P\left( {{A \over {B'}}} \right)

=

{{P\left( {A \cap B'} \right)} \over {P\left( {B'} \right)}}

=

{{P\left( A \right) - P\left( {A \cap B} \right)} \over {1 - P\left( B \right)}}

=

{{{1 \over 3} - {1 \over {18}}} \over {1 - {1 \over 6}}}

=

{{1 \over 3}}

Q47

In a workshop, there are five machines and the probability of any one of them to be out of service on a day is

{{1 \over 4}}

. If the probability that at most two machines will be out of service on the same day is

{\left( {{3 \over 4}} \right)^3}k

, then k is equal to :

A

{{{17} \over 4}}

B

{{{17} \over 2}}

C

{{{17} \over 8}}

D 4

Correct Answer

Option C

Solution

Probablity of at most two machines will be out of service =

{\left( {{3 \over 4}} \right)^3}k

$\Rightarrow$ 5C0

{\left( {{3 \over 4}} \right)^5}

+ 5C1

\left( {{1 \over 4}} \right){\left( {{3 \over 4}} \right)^5}

+ 5C2

{\left( {{1 \over 4}} \right)^2}{\left( {{3 \over 4}} \right)^5}

=

{\left( {{3 \over 4}} \right)^3}k

$\Rightarrow$

{{17} \over 8}{\left( {{3 \over 4}} \right)^3}

=

{\left( {{3 \over 4}} \right)^3}k

$\Rightarrow$ k =

{{17} \over 8}

Q48

A fair coin is tossed a fixed number of times. If the probability of getting 7 heads is equal to probability of getting 9 heads, then the probability of getting 2 heads is :

A

{{15} \over {{2^8}}}

B

{{15} \over {{2^{12}}}}

C

{{15} \over {{2^{13}}}}

D

{{15} \over {{2^{14}}}}

Correct Answer

Option C

Solution

Let the coin be tossed n-times

P(H) = P(T) = {1 \over 2}

P(7 heads) =

{}^n{C_7}{\left( {{1 \over 2}} \right)^{n - 7}}{\left( {{1 \over 2}} \right)^7} = {{{}^n{C_7}} \over {{2^n}}}

P(9 heads) =

{}^n{C_9}{\left( {{1 \over 2}} \right)^{n - 9}}{\left( {{1 \over 2}} \right)^9} = {{{}^n{C_9}} \over {{2^n}}}

P(7 heads) = P(9 heads)

{}^n{C_7} = {}^n{C_9} \Rightarrow n = 16

P(2 heads) =

{}^{16}{C_2}{\left( {{1 \over 2}} \right)^{14}}{\left( {{1 \over 2}} \right)^2} = {{15 \times 8} \over {{2^{16}}}}

P(2 heads)

= {{15} \over {{2^{13}}}}

Q49

A pack of cards has one card missing. Two cards are drawn randomly and are found to be spades. The probability that the missing card is not a spade, is :

A

{{39} \over {50}}

B

{{3} \over {4}}

C

{{22} \over {425}}

D

{{52} \over {867}}

Correct Answer

Option A

Solution

Consider the events, E1 = missing card is spade E2 = missing card is not a spade A = Two spade cards are drawn

P\left( {{E_1}} \right) = {1 \over 4}

P\left( {{E_2}} \right) = {3 \over 4}

P\left( {{A \over {{E_1}}}} \right) = {{{}^{12}{C_2}} \over {{}^{51}{C_2}}}

P\left( {{A \over {{E_2}}}} \right) = {{{}^{13}{C_2}} \over {{}^{51}{C_2}}}

P\left( {{{{E_2}} \over A}} \right) = {{P\left( {{A \over {{E_2}}}} \right).P\left( {{E_2}} \right)} \over {P\left( {{A \over {{E_1}}}} \right).P\left( {{E_1}} \right) + P\left( {{A \over {{E_2}}}} \right).P\left( {{E_2}} \right)}}

= {{{{{}^{13}{C_2}} \over {{}^{51}{C_2}}}.{3 \over 4}} \over {{{{}^{12}{C_2}} \over {{}^{51}{C_2}}}.{1 \over 4} + {{{}^{13}{C_2}} \over {{}^{51}{C_2}}}.{3 \over 4}}}

=

{{39} \over {50}}

Q50

An unbiased coin is tossed 5 times. Suppose that a variable X is assigned the value of k when k consecutive heads are obtained for k = 3, 4, 5, otherwise X takes the value -1. Then the expected value of X, is :

A

- {3 \over {16}}

B

- {1 \over 8}

C

{1 \over 8}

D

{3 \over {16}}

Correct Answer

Option C

Solution

Number of ways 3 consecutive heads can appers (1) HHHT_ (2) _THHH (3) THHHT $\therefore$ Probablity of getting 3 consecutive heads =

{2 \over {32}}

+

{2 \over {32}}

+

{1 \over {32}}

=

{5 \over {32}}

Number of ways 4 consecutive heads can appers (1) HHHHT (2) THHHH $\therefore$ Probablity of getting 4 consecutive heads =

{1 \over {32}}

+

{1 \over {32}}

=

{2 \over {32}}

Number of ways 5 consecutive heads can appers (1) HHHHH $\therefore$ Probablity of getting 5 consecutive heads =

{1 \over {32}}

Now Probablity of getting 0, 1, and 2 consecutive heads = 1 -

\left( {{5 \over {32}} + {2 \over {32}} + {1 \over {32}}} \right)

=

{{{24} \over {32}}}

Now, Expectation = (-1) $\times$

{{{24} \over {32}}}

+ 3 $\times$

{{{5} \over {32}}}

+ 4 $\times$

{{{2} \over {32}}}

+ 5 $\times$

{{{1} \over {32}}}

=

{1 \over 8}