(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.
(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) \(p(\text{black ball at most 2 times}) = p(none) + p(once) + p(twice)\)
= \(10p^{3}q^{2} + 5p^{4}q + p^{5}\)
= \(\frac{16000}{59049} \times 5(\frac{4}{9})^{4} (\frac{5}{9}) + (\frac{4}{9})^{5}\)
= \(\frac{16000 + 6400 + 1024}{59049}\)
= \(\frac{23424}{59049}\)
= \(0.3967\)
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