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Find the following

BOX 1 contains 5 white balls and 6 black balls. Box 2 contains 6 white & 4 black balls A box is selected at random and then a ball is chosen at random from the selected Box

(I) What is the probability that the ball chosen will be a white ball?

(II) Glven that the ball chosen is white what is the probability that came from box1?

**Mumbai University > Electronics and Telecommunication Engineering > Sem 5 > Random Signal Analysis

Marks: 10M

Year: May 2016

1 Answer
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(I)There are two events:

  • Box 1 is selected and white ball is chosen
  • Box 2 is selected and white ball is chosen

Let B1 denotes Box 1 and B2 denotes B2 respectively and W denotes White Ball

Let P(A)=P(Box 1 is selected and a white ball drawn)

=P($B_1$ )×P(W/$B_1$ )

=1/2×5/11=5/22

Let P(B)=P(Box 2 is selected and a white ball drawn)

=P($B_2$ )×P(W/B2 )

=1/2×6/10=3/10

∴P(W)=P(getting a white ball)=P(A)+P(B)

=5/22+3/10=116/220=29/55

(II) To find: P(B1/W) *Using Bayes Theorem*

P($B_1$/W)=(P($B_1$ ).P(W/$B_1$ ))/(P(W))

P($B_1$/W)=(P($B_1$ ).P(W/$B_1$ ))/($∑_{i=1}^2 P ({B_i }).P(W/{B_i} ) )$

P($B_1$/W)=(5/11×1/2)/(29/55)=25/58

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