(i) If 10% of the rivets produced by a machine are defective, find the probability that out of 5 randomly chosen rivets at the most two will be defective.

(ii) If x denotes the outcome, when a fair dice is tossed, find the M.G.F. of x and hence find the mean and variance of x.

Solution:

(i) p=10%=0.1

q=1-p=0.9

n=5

$P(X=Y)=^{n}C_{r}p^{r}q^{n-r}$

For at the most two are defective r=0, 1, 2

$P(X=0 \ or \ 1 \ or \ 2)=^{5}C_{0}(0.1)^{0}(0.9)^{5}+^{5}C_{1}(0.1)^{1}(0.9)^{4}+^{5}C_{2}(0.1)^{2}(0.9)^{3}$

$P(X=0 \ or \ 1 \ or \ 2)=0.59049+0.32805+0.0729=0.99144$

(ii) Here X taking values 1, 2, 3, 4, 5, 6 with probability $\cfrac{1}{6}$

$M_{o}(t)=E(e^{tx_{i}})=\sum p_{i}e^{tx_{i}}=\cfrac{1}{6}e^{t}+\cfrac{1}{6}e^{2t}+\cfrac{1}{6}e^{3t}+....+\cfrac{1}{6}e^{6t}=\cfrac{1}{6}(e^{t}+e^{2t}+e^{3t}+....+e^{6t})$

$\mu_{1}'=\left[ \cfrac{d}{dt} M_{o}(t) \right]_{t=0}= \cfrac{1}{6} \left[ e^{t}+2e^{2t}+3e^{3t}+....+6e^{6t} \right]_{t=0}=\cfrac{1}{6} \left[ 1+2+3+4+5+6 \right] = \cfrac{21}{6} = \cfrac{7}{2}$

$\therefore Mean=\cfrac{7}{2}$

$\mu_{2}'=\left[ \cfrac{d^{2}}{dt^{2}} M_{o}(t) \right]_{t=0}= \cfrac{1}{6} \cfrac{d}{dt} \left[ e^{t}+2e^{2t}+3e^{3t}+....+6e^{6t} \right]_{t=0}$

$\mu_{2}'= \cfrac{1}{6} \left[ e^{t}+4e^{2t}+9e^{3t}+....+36e^{6t} \right]_{t=0}=\cfrac{1}{6} \left[ 1+4+9+16+25+36 \right] = \cfrac{1}{6}[96]$

$\mu_{2}'= \cfrac{96}{9}$

Variance=$\mu_{2}'-\mu_{1}'^{2}=\cfrac{96}{9}-\left( \cfrac{7}{2} \right)^{2}=\cfrac{96}{9}-\cfrac{49}{4}=\cfrac{35}{12}$