Probability — Page 19 | JEE Mathematics

Q181

In a class of 60 students, 40 opted for NCC, 30 opted for NSS and 20 opted for both NCC and NSS. If one of these students is selected at random, then the probability that the students selected has opted neither for NCC nor for NSS is :

A

{1 \over 3}

B

{1 \over 6}

C

{2 \over 3}

D

{5 \over 6}

Correct Answer

Option B

Solution

A $\to$ opted NCC B $\to$ opted NSS $\therefore$ P (nither A nor B) $=$

{{10} \over {60}} = z{1 \over 6}

Q182

In a game, a man wins Rs. 100 if he gets 5 or 6 on a throw of a fair die and loses Rs. 50 for getting any other number on the die. If he decides to throw the die either till he gets a five or a six or to a maximum of three throws, then his expected gain/loss (in rupees) is :

A

{{400} \over 3}

loss

B 0

C

{{400} \over 9}

loss

D

{{400} \over 3}

gain

Correct Answer

Option B

Solution

Expected Gain/Loss = = w $\times$ 100 + Lw( $-$ 50 + 100) + L2w ( $-$ 50 $-$ 50 + 100) + L3( $-$ 150) =

{1 \over 3} \times 100 + {2 \over 3}.{1 \over 3}\left( {50} \right) + {\left( {{2 \over 3}} \right)^2}\left( {{1 \over 3}} \right)\left( 0 \right)

+ {\left( {{2 \over 3}} \right)^3}\left( { - 150} \right) = 0

here w denotes probability that outcome 5 or 6 (w =

{2 \over 6} = {1 \over 3}

) here L denotes probability that outcome 1,2,3,4 (L =

{4 \over 6}

=

{2 \over 3}

)

Q183

An experiment succeeds twice as often as it fails. The probability of at least 5 successes in the six trials of this experiment is :

A

{{240} \over {729}}

B

{{192} \over {729}}

C

{{256} \over {729}}

D

{{496} \over {729}}

Correct Answer

Option C

Solution

Probability of fail, P(F) = p $\therefore$ Probability of success, P(S) = 2p We know, p + 2Pp = 1 $\Rightarrow$ p =

{1 \over 3}

$\therefore$ Now, probability of at least 5 success in 6 trials, P(x $\ge$ 5) = P(x = 5) + P (x = 6) = 6C5

{\left( {{2 \over 3}} \right)^5}{\left( {{1 \over 3}} \right)^1}{ + ^6}{C_6}{\left( {{2 \over 3}} \right)^6}{\left( {{1 \over 3}} \right)^o}

=

{\left( {{2 \over 3}} \right)^5}\left( {{6 \over 3} + {2 \over 3}} \right)

=

{{256} \over {729}}

Q184

Two different families A and B are blessed with equal numbe of children. There are 3 tickets to be distributed amongst the children of these families so that no child gets more than one ticket. If the probability that all the tickets go to the children of the family B is

{1 \over {12}},

then the number of children in each family is :

A 3

B 4

C 5

D 6

Correct Answer

Option C

Solution

Let the number of children in each family be x.

Thus the total number of children in both the families are 2x Now, it is given that 3 tickets are distributed amongst the children of these two families.

Thus, the probability that all the three tickets go to the children in family B =

{{{}^x{C_3}} \over {{}^{2x}{C_3}}}

=

{1 \over {12}}

$\Rightarrow$

\,\,\,

{{x\left( {x - 1} \right)\left( {x - 2} \right)} \over {2x\left( {2x - 1} \right)\left( {2x - 2} \right)}}

=

{1 \over {12}}

$\Rightarrow$

{{\left( {x - 2} \right)} \over {\left( {2x - 1} \right)}}

=

{1 \over 6}

Thus, the number of children in each family is 5.

Q185

Two cards are drawn successively with replacement from a well-shuffled deck of 52 cards. Let X denote the random variable of number of aces obtained in the two drawn cards. Then P(X = 1) + P (X = 2) equals :

A

25 \over 169

B

49\over 169

C

24 \over 169

D

52 \over 169

Correct Answer

Option A

Solution

P (X = 1) means out of two drawn cards one card is ace. and P(X = 2) means both the drawn cards are ace.

$\therefore$ P(X = 1) = first card is ace or 2nd card is ace.

= A _ $+$ _ A =

{4 \over {52}} \times {{48} \over {52}} + {{48} \over {52}} \times {4 \over {52}}

=

2 \times {4 \over {52}} \times {{48} \over {52}}

P(X = 2) = First and second both cards arc ace. = A A =

{4 \over {52}} \times {4 \over {52}}

$\therefore$ P(X = 1) + P(X = 2) =

2 \times {4 \over {52}} \times {{48} \over {52}} + {4 \over {52}} \times {4 \over {52}}

=

{{25} \over {169}}

Q186

Events

A, B, C

are mutually exclusive events such that

P\left( A \right) = {{3x + 1} \over 3},

P\left( B \right) = {{1 - x} \over 4}

and

P\left( C \right) = {{1 - 2x} \over 2}

The set of possible values of

x

are in the interval.

A

\left[ {0,1} \right]

B

\left[ {{1 \over 3},{1 \over 2}} \right]

C

\left[ {{1 \over 3},{2 \over 3}} \right]

D

\left[ {{1 \ 3},{13 \over 3}} \right]

Correct Answer

Option B

Solution

Given

P\left( A \right) = {{3x + 1} \over 3},

P\left( B \right) = {{1 - x} \over 4}

and

P\left( C \right) = {{1 - 2x} \over 2}

We know for any event X,

0 \le P\left( X \right) \le 1

$\therefore$

0 \le {{3x + 1} \over 3} \le 1

\Rightarrow - 1 \le 3x \le 2

\Rightarrow - {1 \over 3} \le x \le {2 \over 3}

0 \le {{1 - x} \over 4} \le 1

\Rightarrow - 3 \le x \le 1

0 \le {{1 - 2x} \over 2} \le 1

\Rightarrow - 1 \le 2x \le 1

\Rightarrow - {1 \over 2} \le x \le {1 \over 2}

In the question given that A, B and C are mutually exclusive So

P\left( {A \cup B \cup C} \right)

=

P\left( {A} \right)

+

P\left( {B} \right)

+

P\left( {C} \right)

\Rightarrow P\left( {A \cup B \cup C} \right)

=

{{3x + 1} \over 3}

+

{{1 - x} \over 4}

+

{{1 - 2x} \over 2}

$\therefore$ 0 $\le$

{{3x + 1} \over 3}

+

{{1 - x} \over 4}

+

{{1 - 2x} \over 2}

$\le$ 1 0 $\le$ 13 - 3

x

$\le$ 12

\Rightarrow {1 \over 3} \le x \le {{13} \over 3}

From all those relations, we get

\max \left\{ { - {1 \over 3}, - 3, - {1 \over 2},{1 \over 3}} \right\}

$\le$

x

$\le$

\min \left\{ {{2 \over 3},1,{1 \over 2},{{13} \over 3}} \right\}

So,

{1 \over 3} \le x \le {1 \over 2}

$\Rightarrow$

\Rightarrow x \in \left[ {{1 \over 3},{1 \over 2}} \right]