Question

Probability of selecting in company a is 1/4. probability of selecting in company b is 1/6. what is the probability of getting selected in both and probability of not getting selected in both?

Answer 1

1/4+1/6=6+4/24=10/24=5/12

Answer 2

Answer:

1) $P(A\cap B)=\frac{1}{24}$

2) $P(A\cup B)=\frac{3}{8}$

Step-by-step explanation:

Given : Probability of selecting in company a is $\frac{1}{4}$.

Probability of selecting in company b is  $\frac{1}{6}$.

To find : What is the probability of getting selected in both and probability of not getting selected in both?

Solution :

Probability of selecting in company a is $P(A)=\frac{1}{4}$

Probability of selecting in company b is  $P(B)=\frac{1}{6}$

1) The probability of getting selected in both is given by,

$P(A\text{ and } B)=P(A)\times P(B)$

$P(A\cap B)=\frac{1}{4}\times \frac{1}{6}$

$P(A\cap B)=\frac{1}{24}$

2) The probability of not getting selected in both is given by,

$P(A\cup B)=P(A)+P(B)-P(A\cap B)$

$P(A\cup B)=\frac{1}{4}+\frac{1}{6}-\frac{1}{24}$

$P(A\cup B)=\frac{6+4-1}{24}$

$P(A\cup B)=\frac{9}{24}$

$P(A\cup B)=\frac{3}{8}$