Question

Abby, Billy and Cathy fire one shot each at a target. The probability that Abby will hit the target is 1/5. The probability that Billy will hit the target is 1/4. The probability that Cathy will hit the target is 1/3.
If they fire together, calculate the probability that:

1. All three shots hit the target

2. Only Cathy's shot hits the target

3. At least one shot hits the target

Answer 1

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

$\displaystyle P(A\bigcap B)= \ P(A).P(B)$

Being P(A) and P(B) the individual probability of each independent event

The probability that A does not occur is  

$\displaystyle P(\bar{A})=1-P(A)$

The probability that B does not occur is

$\displaystyle P(\bar{B})=1-P(B)$

The probability that C does not occur is

$\displaystyle P(\bar{C})=1-P(C)$

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.  

$\displaystyle P(A)=\frac{1}{5}$

The probability that Billy will hit the target is 1/4.

$\displaystyle P(B)=\frac{1}{4}$

The probability that Cathy will hit the target is 1/3

$\displaystyle P(C)=\frac{1}{3}$

Part 1.

The probability that all three shots hit the target is

$\displaystyle P(A\bigcap B\bigcap C)=\frac{1}{5}.\frac{1}{4}.\frac{1}{3}$

$\displaystyle P(A\bigcap B\bigcap C)=\frac{1}{60}$

The probability that all three shots hit the target is

$\displaystyle P= \frac{1}{60}=0.0167$

Part 2.

The probability that only Cathy's shot hits the target is computed assuming Abby and Billy won't succeed

.

$\displaystyle P(\bar{A}\bigcap \bar{B}\bigcap C)=P(\bar{A})\ P(\bar{B})\ P(C)=(1-\frac{1}{5})\ (1-\frac{1}{4})\ (\frac{1}{3})=\frac{4}{5}.\frac{3}{4}.\frac{1}{3}=\frac{1}{5}=0.2$

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

$\displaystyle P(A\bigcap \bar{B}\bigcap \bar{C})+P(\bar{A}\bigcap {B}\bigcap \bar{C})+P(\bar{A}\bigcap\bar{B}\bigcap C)+P(A\bigcap B\bigcap \bar{C})+P(A\bigcap \bar{B}\bigcap C)+P(\bar{A}\bigcap B\bigcap C)+P(A\bigcap B\bigcap C)$

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

$\displaystyle P=1-P(\bar{A}\bigcap \bar{B}\bigcap \bar{C})$

$P=1 -(1-\frac{1}{5})(1-\frac{1}{4})(1-\frac{1}{3})=1-\frac{4}{5}.\frac{3}{4}.\frac{2}{3}=1-\frac{2}{5}=\frac{3}{5}$

P=0.6