Answer:
1. 0.0167
2. 0.2
3. 0.6
Step-by-step explanation:
Probabilities on Independent Events
If A and B are independent events (the occurrence of A doesn't affect the occurrence of B and vice-versa), then the probability that both events occur is
[tex]\displaystyle P(A\bigcap B)= \ P(A).P(B)[/tex]
Being P(A) and P(B) the individual probability of each independent event
The probability that A does not occur is
[tex]\displaystyle P(\bar{A})=1-P(A)[/tex]
The probability that B does not occur is
[tex]\displaystyle P(\bar{B})=1-P(B)[/tex]
The probability that C does not occur is
[tex]\displaystyle P(\bar{C})=1-P(C)[/tex]
We have 3 independent events. We know that because they fire together, no mutual affectation can happen
.
The probability that Abby will hit the target is 1/5.
[tex]\displaystyle P(A)=\frac{1}{5}[/tex]
The probability that Billy will hit the target is 1/4.
[tex]\displaystyle P(B)=\frac{1}{4}[/tex]
The probability that Cathy will hit the target is 1/3
[tex]\displaystyle P(C)=\frac{1}{3}[/tex]
Part 1.
The probability that all three shots hit the target is
[tex]\displaystyle P(A\bigcap B\bigcap C)=\frac{1}{5}.\frac{1}{4}.\frac{1}{3}[/tex]
[tex]\displaystyle P(A\bigcap B\bigcap C)=\frac{1}{60}[/tex]
The probability that all three shots hit the target is
[tex]\displaystyle P= \frac{1}{60}=0.0167[/tex]
Part 2.
The probability that only Cathy's shot hits the target is computed assuming Abby and Billy won't succeed
.
[tex]\displaystyle P(\bar{A}\bigcap \bar{B}\bigcap C)=P(\bar{A})\ P(\bar{B})\ P(C)=(1-\frac{1}{5})\ (1-\frac{1}{4})\ (\frac{1}{3})=\frac{4}{5}.\frac{3}{4}.\frac{1}{3}=\frac{1}{5}=0.2[/tex]
3.
The probability that at least one shot hits the target is when one of them succeeds, two of them succed or all of them succeed
[tex]\displaystyle P(A\bigcap \bar{B}\bigcap \bar{C})+P(\bar{A}\bigcap {B}\bigcap \bar{C})+P(\bar{A}\bigcap\bar{B}\bigcap C)+P(A\bigcap B\bigcap \bar{C})+P(A\bigcap \bar{B}\bigcap C)+P(\bar{A}\bigcap B\bigcap C)+P(A\bigcap B\bigcap C)[/tex]
But it's easier to find the negated probability of the above, i.e. we compute the probability that NO ONE hits the target and subtract it from 1
[tex]\displaystyle P=1-P(\bar{A}\bigcap \bar{B}\bigcap \bar{C})[/tex]
[tex]P=1 -(1-\frac{1}{5})(1-\frac{1}{4})(1-\frac{1}{3})=1-\frac{4}{5}.\frac{3}{4}.\frac{2}{3}=1-\frac{2}{5}=\frac{3}{5}[/tex]
P=0.6
