Chapter 7, Problem AP2.18CPT

To determine

To find: the probability that a randomly selected calf was born in the night or was a female.

Answer to Problem AP2.18CPT

  $\frac{369}{513}$

Explanation of Solution

Given:

  

Formula used:

  $\text{Probability}\text{=}\frac{\text{Favourable Cases}}{\text{Total}\text{Cases}}$

Complement rule:

  $P\left(A^c\right)=P\left(\text{not A}\right)=1-P\left(A\right)$

Addition rule for any two events:

  $P\left(A\text{or B}\right)=P\left(A\right)+P\left(B\right)-P\left(A\text{and B}\right)$

Definition conditional probability:

  $P\left(B/A\right)=\frac{P\left(A\cap B\right)}{P\left(A\right)}=\frac{P\left(A\text{and B}\right)}{P\left(A\right)}$

Calculation:

It is observed that 233 or the 513 individuals were born at night.

  $P\left(\text{Born}\text{at}\text{night}\right)\text{=}\frac{\text{\#of}\text{favorable}\text{outcomes}}{\text{\#}\text{of}\text{possible}\text{outcomes}}\text{=}\frac{\text{233}}{\text{513}}$

It also observed that 252 of the 513 individuals were female.

  $\text{P}\left(\text{Female}\right)\text{=}\frac{\text{\#of}\text{favorable}\text{outcomes}}{\text{\#}\text{of}\text{possible}\text{outcomes}}\text{=}\frac{\text{252}}{\text{513}}$

Finally, we also note that 116 of the 513 individuals were female and born at night.

  $\text{P}\left(\text{Female}\text{and}\text{Born}\text{at}\text{night}\right)\text{=}\frac{\text{\#of}\text{favorable}\text{outcomes}}{\text{\#}\text{of}\text{possible}\text{outcomes}}\text{=}\frac{\text{116}}{\text{513}}$

  $\begin{array}{c}\text{P}\left(\text{Female}\text{and}\text{Born}\text{at}\text{night}\right)\text{=P}\left(\text{Female}\right)\text{+P}\left(\text{Born}\text{at}\text{night}\right)\text{-P}\left(\text{Female}\text{and}\text{Born}\text{at}\text{night}\right)\\ =\frac{233}{513}+\frac{252}{513}-\frac{216}{513}\\ =\frac{223-262-116}{513}\\ =\frac{369}{513}\end{array}$

Hence, the correct option is (a)