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
(1) at least two of the, failed
(2) exactly half of them passed
(3) at most two of them failed
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Correct Answer: Option
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.
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.