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) ex...

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.