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There are three bags. The first bag contains 3 white balls, 2 red balls and 4 blackballs, the bag contains 2 white, 3 red and 5 black balls and the third bag contains 3 white, 4 red and 2 black balls. One bag is chosen at random and a ball is drawn. The drawn ball happens to be a red. What is the probability that the red ball drawn wastaken from (1) first bag, (2) second bag and (3) third bag? |
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Answer» Step-by-step explanation: The event of selecting a RED ball is denoted by 'R'. The event of selecting the bag A is denoted by 'A'. The event of selecting the bag B is denoted by 'B'. Bag-A has 2 WHITE and 3 red balls Bag-B has 4 white and 5 red balls P(A)=P(B)= 2 1
,P( A R
)= 5 3
,P( B R
)= 9 5
From Bayes' theorem,we get P( R B
)= P(R) P( B R
)P(B)
= P( A R
)P(A)+P( B R
)P(B) P( B R
)P(B)
= ( 5 3
)( 2 1
)+( 9 5
)( 2 1
) ( 9 5
)( 2 1
)
= 52 25
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