Suppose you have a hat containing ten balls numbered from 0

Suppose you have a hat containing ten balls numbered from 0 to 9. You draw one ball at random and discover that it numbered 7

If you place the ball back into the hat, what is the probability that you pull out the 7 ball on your next draw?

A) 0.5 (either you draw a 7 or you don’t)       B) 1/10       C) 1/9       D) 0       E) 1

If you place the ball back into the hat, what is the probability that you pull out the 3 ball on your next draw?

A) 0.5 (either you draw a 3 or you don’t)       B) 1/10       C) 1/9       D) 0       E) 1

If you do not place the ball back into the hat, what is the probability that you pull out the 7 ball on your next draw?

A) 0.5 (either you draw a 7 or you don’t)       B) 1/10       C) 1/9       D) 0       E) 1

If you do not place the ball back into the hat, what is the probability that you pull out the 3 ball on your next draw?

A) 0.5 (either you draw a 3 or you don’t)       B) 1/10       C) 1/9       D) 0       E) 1

If you do not place the ball back into the hat, what is the probability that you pull out even-numbered ball on your next draw?

A) 1/10       B) 1/9       C) 5/9       D) 5/10       E) 0

Solution

One ball out of 10 can be drawn in 10c1=10. (a):By replacing, If we pull out the 7 ball on our draw=1c1, Required probability=1c1/10=1/10. (b): By replacing,If we pull out the 3 ball on our draw=1c1, Required probability=1c1/10=1/10.

Without replacing again we draw one ball out of 9 can be drawn in 9c1=9 (c):Without replacing, If we pull out the 7 ball on our draw=1c1, Required probability=1c1/9=1/9. (d): Without replacing,If we pull out the 3 ball on our draw=1c1, Required probability=1c1/9=1/9.

(e):Cases are favourable to get even-numbered ball is (0,2,4,6,8) out of 9 balls only i.e., 5c1.

Required probability=5/9.