Question

A bag contains 3 red and 4 black balls and another bag has 4 red, 2 black and 3 green balls. The selection of the two bags are equiprobable. A bag is selected at random. From the selected bag a ball is taken out. The drawing of each ball is also equiprobable.
A is the event: “the first bag is selected”
B is the event: “the second bag is selected”
H is the event: “the ball drawn is red”
Then find P(A), P(B), P(H/A) and P(H/B).

Answer 1

Answer on Question #42089 - Math - Statistics and Probability

A bag contains 3 red and 4 black balls and another bag has 4 red, 2 black and 3 green balls. The selection of the two bags are equiprobable. A bag is selected at random. From the selected bag a ball is taken out. The drawing of each ball is also equiprobable.

$A$ is the event: "the first bag is selected"

$B$ is the event: "the second bag is selected"

$H$ is the event: "the ball drawn is red"

Then find $P(A), P(B), P(H|A)$ and $P(H|B)$.

Solution.

The selection of the two bags are equiprobable, so $P(A) = \frac{1}{2} = 0.5$.

Similarly, $P(B) = \frac{1}{2} = 0.5$.

$P(H|A)$ is the probability of $H$ provided first bag has already chosen. So, $P(H|A) = \frac{m}{n}$, $m$ is amount of red balls in the bag, $n$ is amount of balls in the bag. Then, $P(H|A) = \frac{3}{7}$.

Similarly, $P(H|B) = \frac{4}{9}$.

Answer: $P(A) = P(B) = \frac{1}{2}, P(H|A) = \frac{3}{7}, P(H|B) = \frac{4}{9}$.

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