In an examination, 60% of the candidates passed. If 10 candidates are selected at random, find the probability that;
(1) at least two of the, failed
(2) exactly half of them passed
(3) at most two of them failed
Explanation
The probability of success (p) = (frac{3}{5}) and the probability of failure (q) = 1 – (frac{3}{5} = frac{2}{5})
The probability that at least two of them failed is
P(x (geq) 2) = 1 – p(x < 2) = 1 – [P(x = 0)] + [P(x = 1)]= 1 – [((^{10}_{0})) ((frac{2}{5}))(^0) ((frac{3}{5}))(^{10}) + ((^{10}_1)) ((frac{2}{5}))(^1) ((frac{3}{5}))(^9)]
Which simplified to 1 – 0.046357401 = 0.9536 correct to four decimal places
(b) The probability that exactly half of them passes is P(x = 6) = ((^{10}_{5})) ((frac{3}{5}))(^5) ((frac{2}{5}))(^{5})
Simplified to give; 252 x 0.7776 x 0.01024 = 0.2007 correct to four decimal places
(c) The probability that at most two of them failed is
P(x (leq) 2) = P(x = 0) + P(x = 1) + P(x = 2). In this case p = (frac{2}{5}) q = (frac{3}{5}) so that
p(x (leq) 2) = ((^{10}_{0})) ((frac{2}{5}))(^0) ((frac{3}{5}))(^{10}) + ((^{10}_1)) ((frac{2}{5}))(^1) ((frac{3}{5}))(^9) + ((^{10}_{2})) ((frac{2}{5}))(^2) ((frac{3}{5}))(^{8})
Which simplies to 0.1673 correct to four decimal places.