A bag contains 10 blue, 6 green, and 4 red marbles. You choose one marble. Without putting it back, you choose a second marble. What is the probability that you first choose a green marble and then a blue marble?
Answer:
\(\frac{3}{19}\)
Step-by-step explanation:
Given :
A bag contains 10 blue, 6 green, and 4 red marbles.
You choose one marble. Without putting it back, you choose a second marble.
To Find: What is the probability that you first choose a green marble and then a blue marble?
Solution:
No. of blue balls = 10
No. of green balls =6
No. of red marbles =4
Total no. of marbles = 10+6+4 =20
Since we are given that first he choose a green marble
So, probability of getting green marble :
\(\frac{\text{No. of green marbles}}{\text{total no. of marbles}}\)
= \(\frac{6}{20}\)
Since he choose second marble without replacement
So, after choosing first ball . The total no. of balls will be 19
So, Probability of getting blue marble in second draw :
\(\frac{\text{No. of blue marbles}}{\text{total no. of marbles}}\)
= \(\frac{10}{19}\)
The probability that you first choose a green marble and then a blue marble:
\(=\frac{6}{20}*\frac{10}{19}\)
\(=\frac{3}{19}\)
Hence the probability that you first choose a green marble and then a blue marble is \(\frac{3}{19}\)