Question

Let’s play Pick-A-Ball with replacement! There are 10 colored balls: 3 red, 4 white, and 3 blue. The balls have been placed into a small bucket, and the bucket has been shaken thoroughly. You will be asked to reach into the bucket, without looking, and select two balls. Since the bucket has been shaken thoroughly, you can assume that each individual ball is selected at random with equal likelihood of being chosen. Now, close your eyes! Reach into the bucket, and pick a ball. (Click "Pick-A-Ball!" to select your ball.) Pick-A-Ball! What is the probability of selecting the color of ball that you just selected? (Enter your answer in decimal format and round it to two decimal places.)

Answer 1

Answer:

0.48

Step-by-step explanation:

There are 10 colored balls: 3 red, 4 white, and 3 blue.

You selected 2 balls at random. They may be

RR, WW, BB, RW, RB, WB, WR, BR, BW.

To find the probability of selecting the color of ball that you just selected, find this probability in each of previous cases:

RR: (One red ball left and 8 balls left in total)

$P_{RR}=\dfrac{3}{10}\cdot \dfrac{2}{9}\cdot \dfrac{1}{8}=\dfrac{1}{120}$

WW: (Two white balls left and 8 balls left in total)

$P_{WW}=\dfrac{4}{10}\cdot \dfrac{3}{9}\cdot \dfrac{2}{8}=\dfrac{1}{30}$

BB: (One blue ball left and 8 balls left in total)

$P_{BB}=\dfrac{3}{10}\cdot \dfrac{2}{9}\cdot \dfrac{1}{8}=\dfrac{1}{120}$

RW: (Two red and three white balls left and 8 balls left in total)

$P_{RW}=\dfrac{3}{10}\cdot \dfrac{4}{9}\cdot \dfrac{5}{8}=\dfrac{1}{12}$

RB: (Two red and two blue balls left and 8 balls left in total)

$P_{RB}=\dfrac{3}{10}\cdot \dfrac{3}{9}\cdot \dfrac{4}{8}=\dfrac{1}{20}$

WB: (Two blue and three white balls left and 8 balls left in total)

$P_{WB}=\dfrac{4}{10}\cdot \dfrac{3}{9}\cdot \dfrac{5}{8}=\dfrac{1}{12}$

WR: (Two red and three white balls left and 8 balls left in total)

$P_{WR}=\dfrac{4}{10}\cdot \dfrac{3}{9}\cdot \dfrac{5}{8}=\dfrac{1}{12}$

BR: (Two red and two blue balls left and 8 balls left in total)

$P_{BR}=\dfrac{3}{10}\cdot \dfrac{3}{9}\cdot \dfrac{4}{8}=\dfrac{1}{20}$

BW: (Two blue and three white balls left and 8 balls left in total)

$P_{BW}=\dfrac{4}{10}\cdot \dfrac{3}{9}\cdot \dfrac{5}{8}=\dfrac{1}{12}$

In total, the probability of selecting the color of ball that you just selected is

$\dfrac{1}{120}+\dfrac{1}{30}+\dfrac{1}{120}+2\cdot\dfrac{1}{12}+2\cdot \dfrac{1}{20}+2\cdot \dfrac{1}{12}=\\ \\=\dfrac{1}{120}+\dfrac{4}{120}+\dfrac{1}{120}+\dfrac{20}{120}+\dfrac{12}{120}+\dfrac{20}{120}=\dfrac{58}{120}=\dfrac{29}{60}\approx 0.48$