A probability question.

Question

A jar contains 3 blue balls, 2 white balls and 5 red balls. A ball is selected at random and the colour of the ball is noted and replaced back into the jar. A second ball is selected and its colour is noted and replaced back into the jar. A third ball is selected and its colour is noted.

what is the probability of selecting 3 red balls, selecting a blue ball, then a white ball, then a blue ball, and what is the probability of selecting a red, then white, then blue ball.

Answer 1

R = red ball

W = white ball

B = blue ball

P(R₁R₂R₃) = (5/10)*(4/9)*(3/8) = 1/12 = 0.0833

P(B₁W₂B₃) = (3/10)*(2/9)*(2/8) = 1/60 = 0.0167

P(R₁W₂B₃) = (5/10)*(2/9)*(3/8) = 1/24 = 0.0417

The probability of selecting 3 red balls is 0.0833.

The probability of selecting a blue ball, then a white ball, then a blue ball is 0.0167.

The probability of selecting a red ball, then a white ball, then a blue ball is 0.0417.

Answer 2

The probability of any event happening depends upon the total number of possible outcomes and the desired outcomes. It is also important to know if items are replaced. In this case, the total number of items is clear: 3 + 2 + 5 = 10 total items. Since each ball that is chosen is replaced, the total number of balls does not change for each selection. To find the probability of each situation, then, simply multiply the probability of each choice together.

a) 3 red balls: 5/10 * 5/10 * 5/10 = 1/2 * 1/2 * 1/2 = 1/8 or 0.125

b) Blue/White/Blue: 3/10 * 2/10 * 3/10 = 18/1000 = 9/500 or 0.018

c) Red/White/Blue: 5/10 * 2/10 * 3/10 = 30/1000 = 3/100 or 0.03

Answer 3

There are 10 total balls and you are sampling with replacement, so each sampling is an independent event..

Probability of selecting a red ball each time is 1/2, so the probability of selecting it three times in a row is 1/2 * 1/2 * 1/2 = 1/8

Probability of selecting a blue ball (3/10), a white ball (1/5) and a blue ball (3/10) consecutively is 3/10 * 1/5 * 3/10 = 9/500

Probability of selecting a red ball (1/2), a white ball (1/5) and a blue ball (3/10) consecutively is 1/2 * 1/5 * 3/10 = 3/300

Answer 4

Sharifah

This is an example of selection with replacement, so probabilities used will not change throughout the problem.

There are 5 red balls out of 10. So the probability of 3 red is p (red) x p (red) x p (red)

p(red) = 5/10. So p (RRR) = 5/10 . 5/10 . 5/10 = 1/2 . 1/2 . 1.2 = 1/8 (simplify the fraction as much as you can).

Then you also need p (BWB) and p (RWB)

p(BWB) = p(B).p(W).p(B).

p(B) = 3/10 (3 BLUE BALLS), and p (W) = 2/10.

I leave the 2nd and 3rd for you as an exercise, using my hints here.

Mike

Answer 5

There appear to be 3 questions:

1. P(3 red balls) = ?
2. P(blue, white, blue) = ?
3. P(red, blue, white) = ?

Each probability can be rewritten as the product of 3 different probabilities:

1. P(3 red balls) = P(red) * P(red) * P(red)
2. P(blue, white, blue) = P(blue) * P(white) * P(blue)
3. P(red, blue, white) = P(red) * P(blue) * P(white)

Note that the order does not matter, since each drawing occurs with replacement (i.e., we are drawing from the same pool of balls each time). For each probability, our denominator is the total number of balls in the bag when a ball is drawn. Again, because we are drawing with replacement, the denominator is always the same:

1. Total balls = 3 + 2 + 5 = 10

The numerator for each probability is the number of balls of the target color that are in the bag. As with the denominator, this value is also constant since we replace each ball we draw. Thus, we have:

1. P(3 red balls) = (5/10) * (5/10) * (5/10) = 1/8
2. P(blue, white, blue) = (3/10) * (2/10) * (3/10) = 9/500
3. P(red, blue, white) = (5/10) * (2/10) * (3/10) = 3/100