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If the independent probabilities that three people $\mathrm{A}, \mathrm{B}$ and $\mathrm{C}$ will be alive in 30 years time are $0.4,0.3,0.2$ respectively, calculate the probability that in 30 years' time,

(a) all will be alive
(b) none will be alive
(c) only one will be alive
(d) at least one will be alive
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(a) $P=P(A) \times P(B) \times P(C)=0.4 \times 0.3 \times 0.2=0.024$
(b) We use the notation $P(\bar{A})$ to mean "the probability that $A$ will not occur". So:
\begin{aligned} &P=P(\bar{A}) \times P(\bar{B}) \times P(\bar{C}) \\ &=0.6 \times 0.7 \times 0.8 \\ &=0.336 \end{aligned}
(c) $P=P(A$ only alive $)+P(B$ only alive $)+P(C$ only alive $)$
\begin{aligned} &=[P(A) \times P(\bar{B}) \times P(\bar{C})]+[P(\bar{A}) \times P(B) \times P(\bar{C})]+[P(\bar{A}) \times P(\bar{B}) \times P(C)] \\ &=0.4 \times 0.7 \times 0.8+0.6 \times 0.3 \times 0.8+0.6 \times 0.7 \times 0.2 \\ &=0.452 \\ &\text { (d) } P=1-\{P(\bar{A}) \times P(\bar{B}) \times P(\bar{C})\}=1-0.336=0.664 \end{aligned}
by Diamond (50,375 points)

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