Question

Find the probability for the experiment of drawing two marbles (without replacement) at random from a bag containing one green, two yellow, and three red marbles.
1. Both marbles are red.
2. Both marbles are yellow.
3. Neither marble is yellow.
4. The marbles are of different colors.

Answer 1

Answer with Step-by-step explanation:

We are given that

Number of  green marbles =1

Number of  yellow marbles =2

Number of red marbles=3

Total number of marbles=1+2+3=6

1.The probability for drawing two marbles of red color  without replacement=$\frac{3}{6}\times \frac{2}{5}=\frac{1}{5}$

By using the formula : Probability,$P(E)=\frac{number\;of\;favorable\;cases}{Total\;number\;of\;cases}$

2.The probability for drawing two marbles of yellow color without replacement=$\frac{2}{6}\times \frac{1}{5}=\frac{1}{15}$

3.The probability for drawing two marbles without replacement when neither marbles is yellow=$1-\frac{1}{15}=\frac{15-1}{15}=\frac{14}{15}$

By using the formula P(E')=1-P(E)

4.The probability for drawing two marbles without replacement when the marbles are of same color=$\frac{1}{15}+\frac{1}{5}=\frac{1+3}{15}=\frac{4}{15}$

The probability for drawing two marbles without replacement when the marbles are of different color=$1-\frac{4}{15}=\frac{15-4}{15}=\frac{11}{15}$