Question

a bag contains 8 red marbles, 7 blue marbles and 2 green marbles. if two marbles are drawn out of the bag (without replacement), what is the exact probability that both marbles drawn will be red?

Explanation:

Step1: Calculate total number of marbles

First, find the total number of marbles in the bag. We have 8 red, 7 blue, and 2 green marbles. So the total number of marbles \( n = 8 + 7 + 2=17 \).

Step2: Calculate number of ways to choose 2 red marbles

The number of ways to choose 2 red marbles out of 8 is given by the combination formula \( C(n,k)=\frac{n!}{k!(n - k)!} \), where \( n = 8 \) and \( k=2 \). So \( C(8,2)=\frac{8!}{2!(8 - 2)!}=\frac{8\times7}{2\times1}=28 \).

Step3: Calculate number of ways to choose 2 marbles from total

The number of ways to choose 2 marbles from 17 is \( C(17,2)=\frac{17!}{2!(17 - 2)!}=\frac{17\times16}{2\times1}=136 \).

Step4: Calculate the probability

The probability that both marbles are red is the number of favorable outcomes (choosing 2 red marbles) divided by the number of total outcomes (choosing 2 marbles from all marbles). So the probability \( P=\frac{C(8,2)}{C(17,2)}=\frac{28}{136}=\frac{7}{34} \).

Answer:

\(\frac{7}{34}\)