Probability — Page 17 | JEE Mathematics

Q161

Minimum number of times a fair coin must be tossed so that the probability of getting at least one head is more than 99% is :

A 6

B 5

C 8

D 7

Correct Answer

Option D

Solution

1 - {\left( {{1 \over 2}} \right)^n} > {{99} \over {100}}

{\left( {{1 \over 2}} \right)^n} < {1 \over {100}}

$\Rightarrow$ n = 7

Q162

Two aeroplanes

{\rm I}

and

{\rm I}

{\rm I}

bomb a target in succession. The probabilities of

{\rm I}

and

{\rm I}

{\rm I}

scoring a hit correctly are

0.3

and

0.2,

respectively. The second plane will bomb only if the first misses the target. The probability that the target is hit by the second plane is :

A

0.2

B

0.7

C

0.06

D 0.32

Correct Answer

Option D

Solution

The desired probability = (0.7) (0.2) + (0.7) (0.8) (0.7) (0.2) + (0.7) (0.8) (0.7) (0.8) (0.7) (0.2) + .......

= 0.14 [1 + (0.56) + (0.56)2 + .......]

= 0.14

\left( {{1 \over {1 - 0.56}}} \right) = {{0.14} \over {0.44}} = {7 \over {22}}

= 0.32

Q163

Let

A

and

B

two events such that

P\left( {\overline {A \cup B} } \right) = {1 \over 6},

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

and

P\left( {\overline A } \right) = {1 \over 4},

where

{\overline A }

stands for complement of event

A

. Then events

A

and

B

are :

A equally likely and mutually exclusive

B equally likely but not independent

C independent but not equally likely

D mutually exclusive and independent

Correct Answer

Option C

Solution

Given that,

P(\overline {A \cup B} ) = {1 \over 6}

,

P(A \cap B) = {1 \over 4}

,

P(\overline A ) = {1 \over 4}

$\because$

P(\overline {A \cup B} ) = {1 \over 6}

\Rightarrow 1 - P(A \cup B) = {1 \over 6}

\Rightarrow 1 - P(A) - P(B) + P(A \cap B) = {1 \over 6}

\Rightarrow P(\overline A ) - P(B) + {1 \over 4} = {1 \over 6}

\Rightarrow P(B) = {1 \over 4} + {1 \over 4} - {1 \over 6}

\Rightarrow P(B) = {1 \over 3}

and

P(A) = {3 \over 4}

Clearly,

P(A \cap B) = P(A)P(B)

, so, the events A and B are independent events but not equally likely.

Q164

If the probability of hitting a target by a shooter, in any shot, is

{1 \over 3}

, then the minimum number of independent shots at the target required by him so that the probability of hitting the target atleast once is greater than

{5 \over 6}

is :

A 4

B 6

C 5

D 3

Correct Answer

Option C

Solution

1 - {}^n{C_0}{\left( {{1 \over 3}} \right)^0}{\left( {{2 \over 3}} \right)^n} > {5 \over 6}

{1 \over 6} > {\left( {{2 \over 3}} \right)^n}\,\, \Rightarrow \,\,0.1666 > {\left( {{2 \over 3}} \right)^n}

{n_{\min }} = 5

Q165

One ticket is selected at random from

50

tickets numbered

00, 01, 02, ...., 49.

Then the probability that the sum of the digits on the selected ticket is

8

, given that the product of these digits is zer, equals :

A

{1 \over 7}

B

{5 \over 14}

C

{1 \over 50}

D

{1 \over 14}

Correct Answer

Option D

Solution

Sample space = {00, 01, 02, 03, ..........49} = 50 tickets n(S) = 50 n(Sum = 8) = { 08, 17, 26, 35, 44 } = 5 n(Product = 0) = { 00, 01, 02, 03, 04, 05, 06, 07, 08, 09, 10, 20, 30, 40 } = 14 $\therefore$ Probability when product is 0 = P(Product = 0) =

{14 \over {50}}

n(Sum = 8 $\cap$ Product = 0) = { 08 } = 1 $\therefore$ Probability when sum is 8 and product is 0 = P(Sum = 8 $\cap$ Product = 0) =

{1 \over {50}}

Required probability,

P\left( {{{Sum = 8} \over {Product = 0}}} \right)

=

{{P\left( {Sum = 8 \cap Product = 0} \right)} \over {P\left( {Product = 0} \right)}}

=

{{{1 \over {50}}} \over {{{14} \over {50}}}}

=

{1 \over {14}}

$\therefore$ Option (D) is correct.

Q166

Let

A

and

B

be two events such that

P\left( {\overline {A \cup B} } \right) = {1 \over 6},\,P\left( { {A \cap B} } \right) = {1 \over 4}

and

P\left( {\overline A } \right) = {1 \over 4},

where

\overline A

stands for the complement of the event

A

. Then the events

A

and

B

are :

A independent but not equally likely.

B independent and equally likely.

C mutually exclusive and independent.

D equally likely but not independent.

Correct Answer

Option A

Solution

P(\overline {A \cup B} ) = {1 \over 6}

or,

1 - P(A \cup B) = {1 \over 6}

$\therefore$

P(A \cup B) = 1 - {1 \over 6} = {5 \over 6};

P(A \cap B) = {1 \over 4}

; and

P(\overline A ) = {1 \over 4};

$\therefore$

P(A) = 1 - P(\overline A ) = 1 - {1 \over 4} = {3 \over 4}

We know,

P(A \cup B) = P(A) + P(B) - P(A \cap B)

or,

{5 \over 6} = {3 \over 4} + P(B) - {1 \over 4}

or,

P(B) = {5 \over 6} - {1 \over 2} = {1 \over 3}

Now,

P(A)\,.\,P(B) = {3 \over 4}.\,{1 \over 3} = {1 \over 4} = P(A \cap B)

i.e., events A and B are mutually independent.

Since the probability of A and B are different, so they are not equally likely events.

Therefore, (A) is the correct option.

Q167

An unbiased coin is tossed eight times. The probability of obtaining at least one head and at least one tail is :

A

{{255} \over {256}}

B

{{127} \over {128}}

C

{{63} \over {64}}

D

{{1} \over {2}}

Correct Answer

Option B

Solution

An unbiased coin is tossed 8 times which is same as 8 coins tossed 1 times.

\therefore\,\,\,

Possible no. of out come = 28

\therefore\,\,\,

Sample space = 28 Here in this condition, all head or all tail out come is not acceptable.

No. of times all head can occur (H H H H H H H H) = 1

\therefore\,\,\,

Probability (all head) =

{1 \over {{2^8}}}

=

{1 \over {256}}

No. of times all tail can occur (T T T T T T T T) = 1

\therefore\,\,\,

Probability (all tail) =

{1 \over {{2^8}}}

=

{{1 \over {256}}}

\therefore\,\,\,

Required probability = 1 $-$ (P (All head) + P (All tail)) = 1 $-$ (

{{1 \over {256}}}

+

{{1 \over {256}}}

) = 1 $-$

{{1 \over {128}}}

=

{{127} \over {128}}

Q168

A random variable

X

has Poisson distribution with mean

2

. Then

P\left( {X > 1.5} \right)

equals :

A

{2 \over {{e^2}}}

B

0

C

1 - {3 \over {{e^2}}}

D

{3 \over {{e^2}}}

Correct Answer

Option C

Solution

In a position distribution,

P(X = r) = {{{e^{ - \lambda }}{\lambda ^r}} \over {r!}}

( $\lambda$ = mean). Now,

P(X = r > 1.5) = P(2) + P(3) +

..... $\infty$

= 1 - \{ P(0) + P(1)\}

= 1 - \left( {{e^{ - 2}} + {{{e^{ - 2}} \times 2} \over 1}} \right) = 1 - {3 \over {{e^2}}}

Q169

A box contains 15 green and 10 yellow balls. If 10 balls are randomly drawn, one-by-one, with replacement, then the variance of the number of green balls drawn is :

A 6

B 4

C

{6 \over {25}}

D

{{12} \over 5}

Correct Answer

Option D

Solution

We can apply binomial probability distribution n = 10 p = Probability of drawing a green ball =

{{15} \over {25}}

=

{3 \over 5}

Also q = 1 -

{3 \over 5}

=

{2 \over 5}

Variance = npq =

10 \times {3 \over 5} \times {2 \over 5}

=

{{12} \over 5}

Q170

A player X has a biased coin whose probability of showing heads is p and a player Y has a fair coin. They start playing a game with their own coins and play alternately. The player who throws a head first is a winner. If X starts the game, and the probability of winning the game by both the players is equal, then the value of 'p' is :

A

{1 \over 5}

B

{1 \over 3}

C

{2 \over 5}

D

{1 \over 4}

Correct Answer

Option B

Solution

P(X getting head) = p $\therefore$ P(X getting tail) = 1 - p P(Y getting head) = P(Y getting tail) =

{1 \over 2}

P(X wins) = p + (1 - p)

{1 \over 2}

p + (1 - p)

{1 \over 2}

(1 - p)

{1 \over 2}

p + ... =

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

=

{{2p} \over {1 + p}}

P(Y win) = (1 - p)

{1 \over 2}

+ (1 - p)

{1 \over 2}

(1 - p)

{1 \over 2}

+ ... =

\left( {{{1 - p} \over 2}} \right).{p \over {1 - \left( {{{1 - p} \over 2}} \right)}} = {{1 - p} \over {1 + p}}

According to question, P(X wins) = P(Y wins) $\therefore$

{{2p} \over {1 + p}}

=

{{1 - p} \over {1 + p}}

$\Rightarrow$ 3p = 1 $\Rightarrow$ p =

{1 \over 3}