ProbabilityHardBloom L3
Question
7 persons are stopped on the road at random and asked about their birthdays. The probability that 3 of them are born on Wednesday, 2 on Thursday and remaining two on Friday is equal to
Options
A.$\frac{141}{7^{7}}$
B.$\frac{45}{7^{7}}$
C.$\frac{90}{7^{6}}$
D.$\frac{30}{7^{6}}$
Solution
Sol. [Key Concept: $7$ persons asked: probability via independent choices.]
$$P = \frac{\frac{7!}{3!2!2!2!} \times 2!}{7^{7}} = \frac{30}{7^{6}}$$
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