GATE Overflow Test Series | Probability | Test 1 | Question: 30

Question

A fair die is thrown as long as necessary for $1$ to turn up. If the number of throws $(n)$ turns out to be even, the probability that $n=4$ is ______

• A. $1/36$
• B. $5/36$
• C. $1/6^4$
• D. None of the above

Answer 1

$P(n = even) = \frac{5}{6}\cdot \frac{1}{6} + \frac{5}{6}\cdot \left (\frac{5}{6} \right )^{2} \cdot \frac{1}{6} + \frac{5}{6}\cdot \left (\frac{5}{6} \right )^{4} \cdot \frac{1}{6} +\frac{5}{6}\cdot \left (\frac{5}{6} \right )^{6} \cdot \frac{1}{6} + ... $

$\implies  P(n=even) =  \frac{\frac{5}{6}\cdot \frac{1}{6}}{1 – \left (\frac{5}{6} \right )^{2} } = \frac{5}{11} $

 

$P({(n=4)}\mid{ (n=even)}) = \frac{ \frac{5}{6}\cdot \left (\frac{5}{6} \right )^{2} \cdot \frac{1}{6}}{\frac{5}{11}} = \frac{5^3\cdot 11}{5\cdot 6^4}  = \frac{5^2\cdot 11}{6^4}$ 

Hence D) none of the above.

Comments

if you mean to ask that if in this same question it would have asked for n=3, instead of n=4,

answer would have been 0, because question is actually asking about conditional probability..

which is  $P(n=3/n=even) = 0$

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Yes, so P(2)+P(4)+P(6)+... should sum to 1. You can see more on this at IISc 👍

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Yes, P(2)+ P(4) + P(6)…  should be 1, (all conditioned on n=even) 

it’ll be as follows

$P(n=even / n=even) = 1$

(after conditioning we reduce our universe to n=even only, and getting any even is ensured by the conditioning as all values in our universe are even only, so conditional probability of getting all evens is 1 )

P(n=odd / n=even) = 0 (because all odds are out of our conditioned universe)

 

One more way we can state the same thing as, sum of all disjoint and mutually exhaustive event in the universe should sum to 1. ( in this case universe has all evens, getting any particular even is disjoint from any other even, and all evens collectively exhaust the whole universe, hence this sum should be 1)

(n=even means all positive even numbers)

 

and thank you Sir, yeah definitely looking forward to grow and learn all sorts of amazing concepts at IISc.

Answer 2

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