(a) Edem and his wife were invited to a dinner by a family of 5.
(a) Edem and his wife were invited to a dinner by a family of 5. They all sat in such a way in such a way that Edem sat next to his wife. Find the number of ways of seating them in a row.
(b) A bag contains 4 red and 5 black identical balls. If 5 balls are selected at random, one after the other with replacement, find the probability that :
(i) a red ball was picked 3 times ; (ii) a black ball was picked at most 2 times.
Explanation
(a) Tie Edem and his wife. There are six people to arrange in 6! ways, Edem and his wife can be arranged in 2! ways.
\(\therefore\) Total number of ways = 2!6! = 1440 ways.
(b) 4 red, 5 black balls
Total = 9 balls.
\(p(red) = p = \frac{4}{9} ; p(black) = q = \frac{5}{9}\)
5 balls are selected. The binomial probability distribution function is
\((p + q)^{5} = p^{5} + 5p^{4}q + 10p^{3}q^{2} + 10p^{2}q^{3} + 5pq^{4} + q^{5}\)
(i) \(p(\text{red ball picked 3 times}) = 10p^{3}q^{2}\)
= \(10 \times (\frac{4}{9})^{3} \times (\frac{5}{9})^{2}\)
= \(10 \times \frac{64}{729} \times \frac{25}{81}\)
= \(0.271\)
(ii) \(p(\text{black ball at most 2 times}) = p(none) + p(once) + p(twice)\)
= \(10p^{3}q^{2} + 5p^{4}q + p^{5}\)
= \(\frac{16000}{59049} \times 5(\frac{4}{9})^{4} (\frac{5}{9}) + (\frac{4}{9})^{5}\)
= \(\frac{16000 + 6400 + 1024}{59049}\)
= \(\frac{23424}{59049}\)
= \(0.3967\)