Test series

Question

i am getting .56

Comments

I am also getting 0.473

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P(R/B1) = P(R ⋂ B1) / P(B1) ==> (3/10) / (1/2)     ???

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Sorry its my mistake it should be 3/10

Answer 1

Bag1(B1) - {3R, 7G}

Bag2(B2) - {4R, 8G}

P(Choosing B1) = 1/2 (if Head)

P(Choosing B2) = 1/2 (if Tail)

 

P(B1/R) = $\frac{P(B1) * P(R/B1)}{P(B1) * P(R/B1) + P(B2) * P(R/B2)}$

$\frac{1/2 * 3/10 * 2/1}{(1/2 * 3/10 * 2/1) + (1/2 * 4/12 * 2/1)}$ = 6/11

 

Point out my mistakes, if any !!

Comments

there is no need of it I think so.

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but why ? can you explain it.

im going acc to method. am i wrong?

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but ur calculation is incorrect: (1/2)(3/10/(1/2))= (1/2)(3*2/10)

if we follow by common sense :

first toss comes head probability =1/2   then occurrence of red ball in bag 1= 3/10

probability of occurrence of red ball from bag one if toss comes head= (1/2)*(3/10)= 3/20