Anonymous
Anonymous asked in 科學及數學數學 · 8 years ago

# Probability

There are 4 black balls, 2 white balls and 1 blue ball in a bag. One ball is drawn at a time at random from the bag without replacement. Find the probability that it takes at least 3 draws to get a black ball.

Rating
• freda
Lv 5
8 years ago

P( takes at least 3 draws)

=P(takes more than or equal to 3 draws)

=P(1- 1 draw -2 draws)

=1- 4/7 -(3/7)*(4/6)

=1-4/7 -2/7

=1/7 //

• 8 years ago

Sol

A：第一次不是黑球

B：第二次不是黑球

C：第三次不是黑球

D：第三是黑球

E：第四次是黑球

P=P(ABCE)+P(ABD)

P(ABCE)

=P(E|ABC)*P(ABC)

=P(E|ABC)*P(C|AB)*P(AB)

=P(E|ABC)*P(C|AB)*P(B|A)*P(A)

=1*(1/5)*(2/6)*(3/7)

P(ABD)

=P(D|AB)*P(AB)

=P(D|AB)*P(B|A)*P(A)

=(4/5)*(2/6)*(3/7)

P=1*(2/6)*(3/7)=1/7

• For it to take at least 3 draws to get a black ball, the first 2 must be non-black ball with probability:

3/7 x 2/6 = 1/7

(Since in the first draw, there are 7 balls in which there are 3 not black and in the 2nd draw, there are 6 balls in which there are 2 not black)

Source(s): 原創答案