Answer:
1) The probability of selecting a white chip is = 21/32
2) The conditional probability that bowl B₁ had been selected, given that a white chip was drawn = 16/21
Step-by-step explanation:
Let
B₁ = The event of randomly selecting Bowl B₁;
B₂ = The event of randomly selecting Bowl B₂ ;
B₃ = The event of randomly selecting Bowl B₃ and
B₄ = The event of randomly selecting Bowl B₄.
The probability of selecting each of the four bowls are as follows, P(B₁) = 1/2, P( B₂ ) = 1/4. P( B₃ ) = 1/8, P( B₄ )=1/8
Let
W = The event of randomly selecting a white chip.
The probability that a white chip is selected from a bowl is given as
P(W | B₁) = 1 for bowl B₁
P(W | B₂ ) = 0 for bowl B₂
P(W | B₃ ) = 1/2 for bowl B₃
P(W | B₄ ) = 3/4 for bowl B₄
There are four ways of selecting a white chip: (1) selecting a white chip from Bowl B₁ ; or (2) selecting a white chip from Bowl B₂; or (3) selecting a white chip from Bowl B₃ or (4) selecting a white chip from Bowl B₄. That is, the probability that a white chip is selected is:
P(W)=P[(W∩B₁)∪(W∩B₂ )∪(W∩B₃) ∪(W∩B₄) ]
Given that the events W∩B₁, W∩B₂ , W∩B₃ and W∩ B₄ are mutually exclusive, and by Multiplication Rule, we have:
P(W)=P(W| B₁ )P(B₁ )+P(W| B₂)P(B₂)+P(W|B₃)P(B₃)+ P(W|B₄)P(B₄)
Substituting the numbers from above
P(W)=(1×1/2)+(0×1/4)+(1/2×1/8)+(3/4×1/8) =1/2+0+1/16+3/32
=21/32
The probability of selecting a white chip is = 21/32
2) P(B₁ Given W), the conditional probability that bowl B₁ had been selected, given that a white chip was drawn.
Solution.
We are interested in finding P(B₁| W). We will use the fact that P(W) = 21/32, as seen from above in our previous calculation:
From conditional probability P(B₁|W) = P(B₁∩W)/(P(W)) = and from multiplication Rule P(W|B₁)×P(B₁)/(P(W)) = 1×(1/2)÷(21/32)= 16/21
Where P(W|B₁) = 1
P(B₁ Given W), the conditional probability that bowl B₁ had been selected, given that a white chip was drawn = 16/21
