(a) Two pupils are chosen at random from a group of 4 boys and 5 girls. Find the probability that the two pupils chosen would be boys.
(b) Twenty percent of the total production of transistors produced by a machine are below standard. If a random sample of 6 transistors produced by the machine is taken, what is the probability of getting (i) exactly 2 standard transistors (ii) exactly 1 standard transistor (iii) at least 2 standard transistors (iv) at most 2 standard transistors?
(a) 4 boys, 5 girls
P(both are boys) = P(1st is a boy and 2nd is a boy)
= \(\frac{4}{9} \times \frac{3}{8}\)
= \(\frac{1}{6}\)
(b) If p is the probability of a standard transistors, then \(p = 0.8 = \frac{4}{5}\)
If q is the probability of a transistor below standard, \(q = 0.2 = \frac{1}{5}\)
For 6 transistors, we have
\((p + q)^{6} = p^{6} + 6p^{5}q + 15p^{4}q^{2} + 20p^{3}q^{3} + 15p^{2}q^{4} + 6pq^{5} + q^{6}\)
(i) p(exactly 2 standard transistors) = \(15p^{2}q^{4}\)
= \(15 \times (\frac{4}{5})^{2} \times (\frac{1}{5})^{4}\)
= \(\frac{48}{3125}\)
(ii) p(exactly 1 standard transistor) = \(6pq^{5}\)
= \(6 \times \frac{4}{5} \times (\frac{1}{5})^{5}\)
= \(\frac{24}{15625}\)
(iii) p(at least 2 standard transistors) = 1 - [p(0 standard transistor) + p(1 standard transistor)]
p(0 standard transistor) = \(q^{6}\)
= \((\frac{1}{5})^{6} = \frac{1}{15625}\)
p(at least 2 standard transistors) = \(1 - [\frac{1}{15625} + \frac{24}{15625}]\)
= \(\frac{15600}{15625} = \frac{624}{625}\)
(iv) p(at most 2 standard transistors) = p(0 standard transistor) + p(1 standard transistor) + p(2 standard transistors)
= \(\frac{1}{15625} + \frac{24}{15625} + \frac{48}{3125}\)
= \(\frac{265}{15625} = \frac{53}{3125}\)
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