In an examination, 70% of the candidates passed. If 12 candidates are selected at random, find the probability that:
(a) at least two of them failed;
(b) exactly half of them passed;
(c) not more than one – six of them failed.
Explanation
70% passed = p
30% failed = q
Using binomial theorem
P[x = x] = \(\begin{pmatrix} n \\ x \end{pmatrix}\)P\(^x\)q\(^{n - x}\)
(a) Probability that at least two of them failed:
P(X \(\geq\) 2) = 1 − P(X=0) − P(X=1)
When P(x = 0) = \(\begin{pmatrix} 12 \\ 12 \end{pmatrix}\)\(\begin{pmatrix} 7 \\ 10 \end{pmatrix}\)\(^2\)\(\begin{pmatrix} 3 \\ 10 \end{pmatrix}\)\(^0\)
= 0.01384129
When P(x = 1) = \(\begin{pmatrix} 12 \\ 11 \end{pmatrix}\)\(\begin{pmatrix} 7 \\ 10 \end{pmatrix}\)\(^{11}\)\(\begin{pmatrix} 3 \\ 10 \end{pmatrix}\)\(^1\)
= 0.07118376
P(X \(\geq\) 2) = 1 - 0.01384129 - 0.0711837 = 0.9149751 ≈ 0.915
(b) half of twelve = 6
P(x = 6) = \(\begin{pmatrix} 12 \\ 6 \end{pmatrix}\) \(\begin{pmatrix} 7 \\ 10 \end{pmatrix}\)\(^6\) \(\begin{pmatrix} 3 \\ 10 \end{pmatrix}\)\(^6\)
= 924 x 0.117649 x 0.000729 = 0.079245
(c) \(\frac{1}{6}\) x 12 = 2, i.e. P[x \(\leq\) 2) = P[x = 0] + P[ x = 1] + P[x = 2]
= 0.01384129 + 0.07118376 + \(\begin{pmatrix} 12 \\ 10 \end{pmatrix}\) \(\begin{pmatrix} 7 \\ 10 \end{pmatrix}\)\(^{10}\) \(\begin{pmatrix} 3 \\ 10 \end{pmatrix}\)\(^2\)
= 0.01384129 + 0.07118376 + 0.167790298 = 0.252815348.