Q. 18 - The Practice of Statistics for AP

Question

Records from a random sample of dairy farms yielded the information below on the number of male and female calves born at various times of the day.

What is the probability that a randomly selected calf was
born in the night or was a female?
(a)$\frac{369}{513}$ 

(b)$\frac{485}{513}$

(c) $\frac{116}{513}$

(d)$\frac{116}{252}$

(e) $\frac{116}{233}$

Step-by-Step Solution

Answer

The correct answer is option (a)$\frac{369}{513}$.

Step 1: Given information

The number of male and female calves born at various times of the day:

	Day	Evening	Night	Total
Males	$129$	$15$	$117$	$261$
Females	$118$	$18$	$116$	$252$
Total	$247$	$33$	$233$	$513$

Step 2: Explanation

The probability rules:
General addition rule:
$P\left(A\text{ or }B\right)=P\left(A\right)+P\left(B\right)-P\left(A\text{ and }B\right)$
Complement rule:
$P\left(A^c\right)=P(notA)=1-P\left(A\right)$
Definition Conditional probability:
$P(B\mid A)=\frac{P(A\cap B)}{P\left(A\right)}=\frac{P\left(A\text{ and }B\right)}{P\left(A\right)}$

Step 3: Calculation

The probability is the number of favorable outcomes divided by the number of possible outcomes.
$P\left(\text{ Born at night }\right)=\frac{233}{513}$
$P\left(\text{ Female }\right)=\frac{262}{513}$
$P\left(\text{ Female and born at night }\right)=\frac{116}{513}$

Hence, the probability that a randomly selected calf was born in the night or female can be determined by:

$$\begin{gathered} P\left(\text{ Female or born at night }\right)=P\left(\text{ Female }\right)+P\left(\text{ Born at night }\right)-P(\text{ Female and born at night)} \\ \begin{array}{r} =\frac{223}{513}+\frac{262}{513}-\frac{116}{513}\end{array} \\ =\frac{223+262-116}{513} \\ =\frac{369}{513} \end{gathered}$$

So, option (a)$\frac{369}{513}$ is correct.