Bowl B1 contains one white and two red chips; bowl B2 contains two white and two red chips; bowl B3 contains one white and four red chips. The probabilities of selecting bowl B1, B2, or B3 are 0.3, 0.2, and 0.5, respectively. A bowl is selected using these probabilities and a chip is then drawn at random. Find: a. P(W), the probability of drawing a white chip.
b. P(B1 Given W), the conditional probability that bowl B1 had been selected, given that a white chip was drawn.

a) 0.3 = 30% probability of drawing a white chip.

b) 0.3333 = 33.33% probability that bowl B1 had been selected, given that a white chip was drawn.

Step-by-step explanation:

Conditional Probability

We use the conditional probability formula to solve this question. It is

In which

P(B|A) is the probability of event B happening, given that A happened.

is the probability of both A and B happening.

P(A) is the probability of A happening.

a. P(W), the probability of drawing a white chip.

1/3(one white out of 3) of 0.3(from B1).

1/2(two white out of 4) of 0.2(from B2).

1/5(one white out of 5) of 0.5(from B3). So

0.3 = 30% probability of drawing a white chip.

b. P(B1 Given W), the conditional probability that bowl B1 had been selected, given that a white chip was drawn.

The probability of drawing a white chip from B1 is 1/3 out of 0.3, so:

Then the conditional probability is:

0.3333 = 33.33% probability that bowl B1 had been selected, given that a white chip was drawn.

it's d.

step-by-step explanation:

because: -1(4x + 1y) +2 (1x-3y) distribute each number into the parenthesis.

then you would get: -4x - 1y + 2x -6y combine like terms

and get: 2x -7y

hope my answer has you if not then i'm sorry.

easy math just plug in a and b

step-by-step explanation: