ISRO2014-37

Question

The probability that two friends are born in the same month is ____ ?

• A. $1/6$
• B. $1/12$
• C. $1/144$
• D. $1/24$

Answer 1

Let us say that what is the probability that they are not born in same month?

A can be born in any of 12 months and B can be born in any of the remaining 11 months $\Rightarrow$ Total possible cases = 12 x 11 = 132

Total Sample space with us : 12 x 12 = 144

$\Rightarrow$ The probability that they are not born in same month = $\frac{132}{144}$

$\Rightarrow$ The probability that they are born in the same month $1 - \frac{132}{144} = \frac{12}{144} \Rightarrow \frac{1}{12}$

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Alternatively,

Let the first friend be born in any of the 12 months. For the second friend we have 12 cases and only 1 is favorable. So, probability = $\frac{12}{12} \times \frac{1}{12} \implies \frac{12}{144} \implies \frac{1}{12}$

Answer 2

$Total\ outcomes = 12*12 =144$

$Favourable\ outcomes = \{$ (Jan,Jan) , (Feb,Feb), (March,March)....(Dec, Dec)$\} = 12$

$Probability = \frac{ Favourable\ outcomes}{Total \ Outcomes} = \frac{12}{12*12} = \frac{1}{12}$

$\therefore$  $B.$ is correct

Comments

Total outcome, why not $12C_2$?

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Since both are Independent events.

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Yes, right.

Answer 3

B) 1/12

Comments

how?

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Desire outcome/ total outcome = probability

total outcome = 12*12 (say if a born in jan then b may born any of the 12 months , if a born in feb then b may born any of the 12 months like wise for 12 months there are 12*12 combination )

Desire outcome = 12 (say if a born in jan then b have to born in jan only one way ,if a born in feb then b have to born in feb only one way, like wise for 12 months there are 12 ways ).peace

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thanks