(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?
Explanation
(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})