Question

1.A bag contains 6 blue balls,4 red balls and 3 yellow balls.
(a.)If a ball is picked at random from the bag,find the probability that it is
red
red and blue
blue or yellow.
2.If two balls are picked at random from the bag (A) with replacement (B)without replacement
Find the probability that
both blue
the first is blue and the second is black
the same colour
different colours

Answer 1

Answer on Question #61607 – Math – Statistics and Probability

Question

1. A bag contains 6 blue balls, 4 red balls and 3 yellow balls.

If a ball is picked at random from the bag, find the probability that it is

red

red and blue

blue or yellow.

Solution

The number of red balls is 4 and the total number of balls is 13, then the probability of picking a red ball is:

$$P = \frac{4}{13}$$

Since the ball cannot be both red and blue, the probability of picking red and blue ball is

$$P = 0$$

The number of blue, yellow balls is 9 and the total number of balls is 13, then the probability of picking a blue or yellow ball is

$$P = \frac{9}{13}$$

Answer: $\frac{4}{13}$; 0; $\frac{9}{13}$.

Question

2. If two balls are picked at random from the bag (A) with replacement (B) without replacement

Find the probability that:

both blue

the first is blue and the second is black

the same colour

different colours

Solution

(A) with replacement.

The number of blue balls is 6 and the total number of balls is 13, then the probability of picking a blue ball is

$$P = \frac{6}{13}$$

Then the probability of picking two blue balls with replacement is

$$P = \frac{6}{13} \cdot \frac{6}{13} = \frac{36}{169}$$

Since the bag does not contain black balls, the probability of picking the first blue and the second black ball is

$$P = 0$$

Let $A_1$ be the event of picking two blue balls. Then probability of $A_1$ is

$$P(A_1) = \frac{36}{169}$$

Let $A_2$ be the event of picking two red balls. Then probability of $A_2$ is

$$P(A_2) = \frac{4}{13} \cdot \frac{4}{13} = \frac{16}{169}$$

Let $A_3$ be the event of picking two yellow balls. Then probability of $A_3$ is

$$P(A_3) = \frac{3}{13} \cdot \frac{3}{13} = \frac{9}{169}$$

Let A be the event of picking two balls of the same color. Then the probability of A is

$$P(A) = P(A_1) + P(A_2) + P(A_3) = \frac{36}{169} + \frac{16}{169} + \frac{9}{169} = \frac{61}{169}$$

Let B be the event of picking two balls of different colors. Then the probability of this event is

$$P(B) = 1 - P(A) = 1 - \frac{61}{169} = \frac{108}{169}$$

(B) without replacement

The number of blue balls is 6 and the total number of balls is 13, then the probability of picking a blue ball is

$$P = \frac{6}{13}$$

After this there are 5 blue balls and totally 12 balls in the bag.

Then the probability of picking two blue balls with replacement is

$$P = \frac{6}{13} \cdot \frac{5}{12} = \frac{30}{156} = \frac{5}{26}$$

Since the bag does not contain black balls, the probability of picking the first blue and the second black ball is

$$P = 0$$

Let $A_1$ be the event of picking two blue balls. Then probability of $A_1$ is

$$P(A_1) = \frac{5}{26}$$

Let $A_2$ be the event of picking two red balls. Then probability of $A_2$ is

$$P(A_2) = \frac{4}{13} \cdot \frac{3}{12} = \frac{12}{156} = \frac{1}{13}$$

Let $A_3$ be the event of picking two yellow balls. Then probability of $A_3$ is

$$P(A_3) = \frac{3}{13} \cdot \frac{2}{12} = \frac{6}{156} = \frac{1}{26}$$

Let A be the event of picking two balls of the same color. Then the probability of picking two balls of the same color is

$$P(A) = P(A_1) + P(A_2) + P(A_3) = \frac{5}{26} + \frac{1}{13} + \frac{1}{26} = \frac{5}{26} + \frac{2}{26} + \frac{1}{26} = \frac{8}{26} = \frac{4}{13}$$

Let B be the event of picking two balls of different colors. Then the probability of this event is

$$P(B) = 1 - P(A) = 1 - \frac{4}{13} = \frac{9}{13}$$

Answer: (A) $\frac{36}{169}$; 0; $\frac{61}{169}$; $\frac{108}{169}$; (B) $\frac{5}{26}$; 0; $\frac{4}{13}$; $\frac{9}{13}$.

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