Question

A house has 7 family members living in it. Then what is the probability that out of those 7 persons, 2 persons were born on the same day of the week?
Options

1080/(7^5)

540/(7^4)

2160/(7^5)

331/(7^7)

Answer 1

Probability will be $\frac{720}{7^6}$

Step-by-step explanation:

The probability that the first person born on a unique day is 1

The second has a probability of $1-\frac{1}{7}=\frac{6}{7}$

The third has a probability of $1-\frac{2}{7}=\frac{5}{7}$

The forth has a probability of $1-\frac{3}{7}=\frac{4}{7}$

The fifth has a probability of $1-\frac{4}{7}=\frac{3}{7}$

The sixth has a probability of $1-\frac{5}{7}=\frac{2}{7}$

The seventh has a probability of $1-\frac{6}{7}=\frac{1}{7}$

So total probability $=1\times \frac{6}{7}\times \frac{5}{7}\times \frac{4}{7}\times \frac{3}{7}\times \frac{2}{7}\times \frac{1}{7}=\frac{720}{7^6}$

Which is not given in any option

Learn more

What is the probability of two persons being born on

the same day (ignoring date)?

(NDA 2008 II)

(a) 1/49

(b) 1/365

(c) 1/7

(d) 2/7

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