Probability — Page 4 | JEE Mathematics

Q31

An unbiased coin is tossed. If the outcome is a head then a pair of unbiased dice is rolled and the sum of the numbers obtained on them is noted. If the toss of the coin results in tail then a card from a well-shuffled pack of nine cards numbered 1, 2, 3, ……, 9 is randomly picked and the number on the card is noted. The probability that the noted number is either 7 or 8 is :

A

{{19} \over {36}}

B

{{15} \over {72}}

C

{{13} \over {36}}

D

{{19} \over {72}}

Correct Answer

Option D

Solution

P\left( A \right) = {1 \over 2} \times {{11} \over {36}} + {1 \over 2} \times {2 \over 9} = {{19} \over {72}}

Q32

Two integers are selected at random from the set {1, 2, ...., 11}. Given that the sum of selected numbers is even, the conditional probability that both the numbers are even is :

A

{2 \over 5}

B

{1 \over 2}

C

{7 \over 10}

D

{3 \over 5}

Correct Answer

Option A

Solution

Since sum of two numbers is even so either both are odd or both are even.

Hence number of elements in reduced samples space = 5C2 + 6C2 So, required probability =

{{{}^5{C_2}} \over {{}^5{C_2} + {}^6{C_2}}}

=

{2 \over 5}

Q33

A bag contains 30 white balls and 10 red balls. 16 balls are drawn one by one randomly from the bag with replacement. If X be the number of white balls drawn, then

\left( {{{mean\,\,of\,X} \over {s\tan dard\,\,deviation\,\,of\,X}}} \right)

is equal to :

A 4

B

3\sqrt 2

C

{{4\sqrt 3 } \over 3}

D

4\sqrt 3

Correct Answer

Option D

Solution

p (probability of getting white ball) =

{{30} \over {40}}

q =

{1 \over 4}

and n = 16 mean = np = 16.

{3 \over 4}

= 12 and standard diviation =

\sqrt {npq}

=

\sqrt {16.{3 \over 4}.{1 \over 4}} = \sqrt 3

$\because$

mean \over standard\,deviation

=

12\over{\sqrt3}

=

4{\sqrt3}

Q34

In a random experiment, a fair die is rolled until two fours are obtained in succession. The probability that the experiment will end in the fifth throw of the die is equal to :

A

{{200} \over {{6^5}}}

B

{{225} \over {{6^5}}}

C

{{150} \over {{6^5}}}

D

{{175} \over {{6^5}}}

Correct Answer

Option D

Solution

\underline {} \,\,\,\underline {} \,\,\,\underline {} \,\,\underline 4 \,\,\underline 4

{1 \over {{6^2}}}\left( {{{{5^3}} \over {{6^3}}} + {{2{C_1}{{.5}^2}} \over {{6^3}}}} \right) = {{175} \over {{6^5}}}

Q35

Let two fair six-faced dice

A

and

B

be thrown simultaneously. If

{E_1}

is the event that die

A

shows up four,

{E_2}

is the event that die

B

shows up two and

{E_3}

is the event that the sum of numbers on both dice is odd, then which of the following statements is

NOT

true?

A

{E_1}

and

{E_2}

are independent.

B

{E_2}

and

{E_3}

are independent.

C

{E_1}

and

{E_3}

are independent.

D

{E_1},

{E_2}

and

{E_3}

are independent.

Correct Answer

Option D

Solution

Total possible outcome with two six faced dice = 62 = 36 When dice A shows up 4, the possible cases are E1 = { (4, 1) (4, 2) (4, 3) (4, 4) (4, 5) (4, 6) } = 6 cases $\therefore$

P\left( {{E_1}} \right) = {6 \over {36}} = {1 \over 6}

When dice B shows up 2, the possible cases are E2 = { (1, 2) (2, 2) (3, 2) (4, 2) (5, 2) (6, 2) } = 6 cases

P\left( {{E_2}} \right) = {6 \over {36}} = {1 \over 6}

{E_1} \cap {E_2}

= Common in both in E1 and E2 = { (4, 2) }

P\left( {{E_1} \cap {E_2}} \right) = {1 \over {36}}

And

P\left( {{E_1}} \right)

.

P\left( {{E_2}} \right)

=

{1 \over 6}

.

{1 \over 6}

=

{1 \over 36}

$\therefore$

P\left( {{E_1} \cap {E_2}} \right)

=

P\left( {{E_1}} \right)

.

P\left( {{E_2}} \right)

$\therefore$ E1 and E2 are independent.

{E_3}

= [ (1, 2), (1, 4), (1, 6), (2, 1), (2, 3), (2, 5), (3, 2), (3, 4), (3, 6), (4, 1), (4, 3), (4, 5), (5, 2), (5, 4), (5, 6), (6, 1), (6, 3), (6, 5) ] = 18 cases

{E_1} \cap {E_3}

= { (4, 1) (4, 3) (4, 5) } = 3 cases $\therefore$

P\left( {{E_1} \cap {E_3}} \right) = {3 \over {36}} = {1 \over {12}}

=

{1 \over 6} \times {1 \over 2}

=

P\left( {{E_1}} \right)

$\times$

P\left( {{E_3}} \right)

$\therefore$ E1 and E3 are independent.

{E_2} \cap {E_3}

= { (1, 2) (3, 2) (5, 2) } = 3 cases $\therefore$

P\left( {{E_2} \cap {E_3}} \right) = {3 \over {36}} = {1 \over {12}}

=

{1 \over 6} \times {1 \over 2}

=

P\left( {{E_2}} \right)

$\times$

P\left( {{E_3}} \right)

$\therefore$ E2 and E3 are independent.

{E_1} \cap {E_2} \cap {E_3}

= 0 $\therefore$

P\left( {{E_1} \cap {E_2} \cap {E_3}} \right)

= 0

P\left( {{E_1}} \right) \times

P\left( {{E_2}} \right) \times

P\left( {{E_3}} \right)

=

{1 \over 6}

$\times$

{1 \over 6}

$\times$

{1 \over 2}

=

{1 \over {72}}

$\therefore$

{E_1},

{E_2}

and

{E_3}

are not independent. $\therefore$ Option (D) is correct.

Q36

Let EC denote the complement of an event E. Let E1 , E2 and E3 be any pairwise independent events with P(E1) > 0 and P(E1

\cap

E2

\cap

E3) = 0. Then P(

E_2^C \cap E_3^C/{E_1}

) is equal to :

A

P\left( {E_3^C} \right)

- P(E2)

B

P\left( {E_2^C} \right)

+ P(E3)

C

P\left( {E_3^C} \right)

-

P\left( {E_2^C} \right)

D P(E3) -

P\left( {E_2^C} \right)

Correct Answer

Option A

Solution

Given E1 , E2 , E3 are pairwise indepedent events so P(E1 $\cap$ E2 ) = P(E1 ).

P(E2 ) and P(E2 $\cap$ E3 ) = P(E2 ).

P(E3 ) and P(E3 $\cap$ E1 ) = P(E3 ).

P(E1 ) and P(E1 $\cap$ E2 $\cap$ E3) = 0 Now

P\left( {{{E_2^C \cap E_3^C} \over {{E_1}}}} \right)

= {{P\left[ {{E_1} \cap \left( {E_2^C \cap E_3^C} \right)} \right]} \over {P\left( {{E_1}} \right)}}

= {{P\left( {{E_1}} \right) - \left[ {P\left( {{E_1} \cap {E_2}} \right) + P\left( {{E_1} \cap {E_3}} \right) - P\left( {{E_1} \cap {E_2} \cap {E_3}} \right)} \right]} \over {P\left( {{E_1}} \right)}}

= {{P\left( {{E_1}} \right) - P\left( {{E_1}} \right).P\left( {{E_2}} \right) - P\left( {{E_1}} \right).P\left( {{E_3}} \right) - 0} \over {P\left( {{E_1}} \right)}}

= 1 - P\left( {{E_2}} \right) - P\left( {{E_3}} \right)

= 1 - P\left( {{E_3}} \right) - P\left( {{E_2}} \right)

= P\left( {E_3^C} \right) - P\left( {{E_2}} \right)

Q37

The probabilities of three events A, B and C are given by P(A) = 0.6, P(B) = 0.4 and P(C) = 0.5. If P(A

\cup

B) = 0.8, P(A

\cap

C) = 0.3, P(A

\cap

B

\cap

C) = 0.2, P(B

\cap

C) =

\beta

and P(A

\cup

B

\cup

C) =

\alpha

, where 0.85

\le \alpha \le

0.95, then

\beta

lies in the interval :

A [0.35, 0.36]

B [0.20, 0.25]

C [0.25, 0.35]

D [0.36, 0.40]

Correct Answer

Option C

Solution

P(A $\cup$ B) = P(A) + P(B) – P(A $\cup$ B) $\Rightarrow$ 0.8 = 0.6 + 0.4 – P(A $\cap$ B) $\Rightarrow$ P(A $\cap$ B) = 0.2 P(A $\cup$ B $\cup$ C) = P(A) + P(B) + P(C) – P(A $\cap$ B) – P(B $\cap$ C) –P(C $\cap$ A) + P(A $\cap$ B $\cap$ C) $\Rightarrow$ $\alpha$ = 0.6 + 0.4 + 0.5 - 0.2 - $\beta$ - 0.3 + 0.2 $\Rightarrow$ $\alpha$ + $\beta$ = 1.2 $\Rightarrow$ $\alpha$ = 1.2 - $\beta$ Given, 0.85

\le \alpha \le

0.95 $\Rightarrow$ 0.85 $\le$ 1.2 - $\beta$ $\le$ 0.95 $\Rightarrow$ 0.25

\le \beta \le

0.35

Q38

Out of 11 consecutive natural numbers if three numbers are selected at random (without repetition), then the probability that they are in A.P. with positive common difference, is :

A

{{10} \over {99}}

B

{{5} \over {33}}

C

{{15} \over {101}}

D

{{5} \over {101}}

Correct Answer

Option B

Solution

Out of 11 consecutive natural numbers either 6 even and 5 odd numbers or 5 even and 6 odd numbers.

Let, E = Even O = Odd Case-1 : E, O, E, O, E, O, E, O, E, O, E 2b = a + c $\Rightarrow$ Even $\Rightarrow$ Both a and c should be either even or odd.

P =

{{{}^6{C_2} + {}^5{C_2}} \over {{}^{11}{C_3}}}

=

{5 \over {33}}

Case -2 : O, E, O, E, O, E, O, E, O, E, O P =

{{{}^5{C_2} + {}^6{C_2}} \over {{}^{11}{C_3}}}

=

{5 \over {33}}

Total probability =

{1 \over 2} \times {5 \over {33}}

+

{1 \over 2} \times {5 \over {33}}

=

{5 \over {33}}

Q39

In a game two players A and B take turns in throwing a pair of fair dice starting with player A and total of scores on the two dice, in each throw is noted. A wins the game if he throws total a of 6 before B throws a total of 7 and B wins the game if he throws a total of 7 before A throws a total of six. The game stops as soon as either of the players wins. The probability of A winning the game is :

A

{5 \over {6}}

B

{5 \over {31}}

C

{31 \over {61}}

D

{30 \over {61}}

Correct Answer

Option D

Solution

Sum total 6 = {(1,5)(2,4)(3,3)(4,2)(5,1)}

P(6) = {5 \over 36}

Sum total 7 = {(1,6)(2,5)(3,4)(4,3)(5,2)(6,1)}

\,P(7) = {6 \over {36}}

=

{1 \over 6}

Game ends and A wins if A throws 6 in 1st throw or A don’t throw 6 in 1st throw, B don’t throw 7 in 1st throw and then A throw 6 in his 2 nd chance and so on.

P(A) = A +

\overline A \overline B A

+

\overline A \overline B \overline A \overline B A

=

{5 \over {36}} + \left( {{{31} \over {36}}} \right)\left( {{{30} \over {36}}} \right)\left( {{5 \over {36}}} \right) +

..... $\infty$ =

{{{5 \over {36}}} \over {1 - \left( {{{31} \over {36}}} \right)\left( {{{30} \over {36}}} \right)}}

=

{{5 \times 36} \over {36 \times 36 - 31 \times 30}}

=

{30 \over {61}}

Q40

The probability that a randomly chosen 5-digit number is made from exactly two digits is :

A

{{150} \over {{{10}^4}}}

B

{{134} \over {{{10}^4}}}

C

{{121} \over {{{10}^4}}}

D

{{135} \over {{{10}^4}}}

Correct Answer

Option D

Solution

Sample space = 9 $\times$ 104 Case - I Out of exactly two digits selected one is zero then favourable cases =

{}^9{C_1}({2^4} - 1)

Case - II Both selected digits are non-zero then favourable cases =

{}^9{C_2}({2^5} - 2)

Probability =

{{9({2^4} - 1) + {{9.8} \over 2}({2^5} - 2)} \over {9 \times {{10}^4}}}

= {{15 + 120} \over {{{10}^4}}} = {{135} \over {{{10}^4}}}