Question

A jar contains black and white marbles. Two marbles are chosen without replacement. The probability of selecting a black marble and then a white marble is 0.40, and the probability of selecting a black marble on the first draw is 0.50. What is the probability of selecting a white marble on the second draw, given that the first marble drawn was black?

Answer 1

probability of selecting a white marble on the second draw = 0.8

Step-by-step explanation:

Let say there are B - Black Marbes

& W White marbles

Total Marbles = B + W

The probability of selecting a black marble and then a white marble is 0.40

P(B) P(W)    = 0.4

P(B) = Selecting Black on First Draw

P (W) = Selecting white on 2nd Draw

probability of selecting a black marble on the first draw is 0.50

=> P(B) = 0.5

=> 0.5 * P(W) = 0.4

=> P(W) = 0.8

=> probability of selecting a white marble on the second draw = 0.8

If a committee has 7 members

https://brainly.in/question/13172461

https://brainly.in/question/13136814

Answer 2

Answer:

0.5

Step-by-step explanation:

let the probability of selecting a black marble be P(B) and that of selecting a white marble be P(W).

The probability of selecting a black marble and then a white marble is equivalent to ;

Probability of selecting a black marble AND probability of selecting a white marble given that a black marble was selected first. That is

P(B) AND P(W/B)= P(B)*P(W/B)=0.4

But P(B)=0.5

Therefore

P(B)*P(W/B)=0.4

0.5*P(W/B)=0.4

P(W/B)=0.8

Probability of selecting a white marble on second draw given that the first marble was black becomes a posterior probability, which is computed as ;

$P(W/B)=\frac{P(W)*(P(B/W)}{P(W)*(P(B/W)+P(W)*P(W/B)}\\ P(W/B)=\frac{0.4}{0.4+0.5*0.8}\\ P(W/B)=0.5$