Question

Catriona C and Daniel D practice archery. Suppose P(C) = 1=4 and P(D) = 1=6 denote their

probabilities of hitting the target; we assume that these probabilities are independent. Find

the probability that:

(i) Catriona does not hit the target.

(ii) Both of them hit the target.

(iii) At least one of them hits the target.

(iv) Neither of them hits the target.

Answer 1

i) $1-P(C)=1-\frac {1} {4}=\frac {3} {4}$

ii)$P(C)*P(D)=\frac {1} {4} * \frac {1} {6}=\frac {1} {24}$

iii)chance that at least one of them hits the target is 1 minus chance that neither of them hits the target

$1-((1-P(C))*(1-P(D)))=1-((1-\frac {1} {4})(1-\frac {1} {6}))=1-\frac {15} {24}=\frac {3} {8}$

iv)$(1-P(C))(1-P(D))=(1-\frac {1} {4})(1-\frac {1} {6})=\frac {5} {8}$