Question

if 16 dates are named at random, What is the probability that 3 of them will be Sundays?

Answer 1

Answer on Question #23926 – Math – Statistics and Probability

Question

If 16 dates are named at random, what is the probability that 3 of them will be Sundays?

Solution

The problem is not well posed, because we are not informed about the set of days: month, year or decade, from which we take days. Suppose we take 16 days from a month that consists of 31 days including 4 Sundays. There are $\binom{31}{16}$ ways to select 16 days and we must choose 3 Sundays out of 4 and $16-3=13$ days from the rest $31-4=27$ days, which are not Sundays.

Thus, by the classical definition of probability, the probability that 3 of them will be Sundays equals

$$\frac {\binom {4} {3} \cdot \binom {27} {13}}{\binom {31} {16}} = \frac {\frac {4 !}{3 ! (4 - 3) !} \cdot \frac {27 !}{13 ! (27 - 13) !}}{\frac {31 !}{16 ! (31 - 16) !}} = \frac {\frac {4 !}{3 ! \cdot 1 !} \cdot \frac {27 !}{13 ! \cdot 14 !}}{\frac {31 !}{16 ! \cdot 15 !}} = \frac {4 !}{3 ! \cdot 1 !} \cdot \frac {27 !}{31 !} \cdot \frac {16 ! \cdot 15 !}{13 ! \cdot 14 !} =$$ $$= \frac {4 \cdot 3 ! \cdot 27 ! \cdot 16 \cdot 15 \cdot 14 \cdot 13 ! \cdot 15 \cdot 14 !}{3 ! \cdot 1 ! \cdot 31 \cdot 30 \cdot 29 \cdot 28 \cdot 27 ! \cdot 13 ! \cdot 14 !} = \frac {4 \cdot 16 \cdot 15 \cdot 14 \cdot 15}{31 \cdot 30 \cdot 29 \cdot 28} =$$ $$= \frac {201600}{755160} = \frac {20160}{75516} = \frac {10080}{37758} = 0.26696.$$

Answer: 0.26696.

Answer provided by https://www.AssignmentExpert.com