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Question

A fair die is tossed repeatedly until a six is obtained. Let X denote the number of tosses required. The conditional probability that X6 given X>3 equals


A

125216

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B

25216

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C

536

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D

2536

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Solution

The correct option is D

2536


Explanation for correct option:

Step 1: Define possible outcomes:

Possible comes =1,2,3,4,5,6

The number of tosses requires to get six on the die =X

Probability of not obtaining six on a fair die =56

Step 2: Find the conditional probability that X6 is given X>3:

As per conditional probability,

PX6X>3=PX6X>3PX>3....1[P(AB)=P(AB)P(B)]

We know,

PX6=P6+P7+P8+......+PX>3=P4+P5+P6+P7+......+

PX6X>3=P6+P7+.....+=PX6

Step 3: Calculate the value of all possible conditions:

PX6=5566+5667+5768+......+

=556611-56 [S=a1-r] where ais the first term and r is the common difference

=556661=5565......2

PX>3=5364+5465+5566+......+

=536411-56=536461=5363......3

Substitute the value of equations 3and2in1 we have,

PX6X>3=55655363

=5262=2536

Hence, Option(D) is the correct answer.


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