Question

A box contains four red balls and eight black balls. Two balls are randomly chosen from the box, and are not
replaced. Let event B be choosing a black ball first and event R be choosing a red ball second.
What are the following probabilities?
P(B) =
P(
RB) =
P(B n R) =
The probability that the first ball chosen is black and the second ball chosen is red is about percent.

Answer 1

P(B) = 8/12

P(R | B) = 4/11

P(B ∩ R) = 8/33

The probability that the first ball chosen is black and the second ball chosen is red is about 24% percent

Solution:

The probability is given as:

$Probability = \frac{\text{number of favorable outcomes}}{\text{total number of possible outcomes}}$

Given that,

A box contains four red balls and eight black balls

Red = 4

Black = 8

Total number of possible outcomes = 12

Let event B be choosing a black ball first and event R be choosing a red ball second.

Find P(B)

$P(B) = \frac{8}{12}$

Find P(B n R)

$P(B n R) = P(B) \times P(R)\\\\P(B n R) = \frac{8}{12} \times \frac{4}{11}\\\\P(B n R) = \frac{8}{33}$

Find P(R | B)

$P(R | B) = \frac{P(R n B)}{P(B)}\\\\P(R | B) = \frac{\frac{8}{33}}{\frac{8}{12}}\\\\P(R | B) = \frac{8}{33} \times \frac{12}{8}\\\\P(R | B) = \frac{4}{11}$

The probability that the first ball chosen is black and the second ball chosen is red is about percent

$\frac{8}{33} \times 100 = 0.24 \times 100 = 24 \%$

Thus the probability that the first ball chosen is black and the second ball chosen is red is about 24% percent

Answer 2

Answer:

A box contains four red balls and eight black balls. Two balls are randomly chosen from the box, and are not replaced. Let event B be choosing a black ball first and event R be choosing a red ball second.

What are the following probabilities?

P(B) =  

✔ 8/12

P(R | B) =  

✔ 4/11

P(B ∩ R) =  

✔ 8/33

The probability that the first ball chosen is black and the second ball chosen is red is about  

✔ 24

percent.

Step-by-step explanation:

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