Bowl b₁ contains two white chips, bowl b₂ contains two red chips, bowl b₃ contains two white chips and two red chips, and bowl b₄ contains three white chips and one red chip. the probabilities of selecting bowl b₁, b₂, b₃, or b₄ are 1/2, 1/4, 1/8 and 1/8, 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.

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

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step-by-step explanation:

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solution:

the given function is ,

      =

the three roots of this quadratic function is , 0 ,8 and 10.all the roots are real and rational.

if a quadratic equation is of the form ,

sum of the roots = a + b +c= 0+8+10=18=  

product of roots taken two at a time = a b + b c+ c a = 0+80+0=80 =  

product of roots = a b c = 0=  

there are 11 doors in a hall starting from 0 to 11.the entry in the hall is permitted through gate number 0 and exit is allowed through only gate number 8 and 10.so, three roots(0,8,10) of quadratic function is given by entry and exit route of this hall.