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Question

Let U1 and U2be two urns such that U1 contains 3 white and 2 red balls and U2 contains only 1 white ball. A fair coin is tossed. If the head appears then 1 ball is drawn at random from U1and put into U2. However, if the tail appears then 2 balls are drawn at random from U1 and put into U2. Now, 1 ball is drawn at random from U2. The probability of the drawn ball from U2 being white is


A

1330

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B

2330

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C

1930

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D

1130

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Solution

The correct option is B

2330


Explanation for the correct options:

Find the required probability:

Urn U1contains 3 white and 2 red balls .

U2 contains only 1 white ball

By using the given condition that is ,

Step 1: The probability of drawing ball from U1, if the ball drawn at random is white.

P1=12×35×22=1×31×5=35

Similarly,

Step 2 : The probability of drawing ball from U2, if the ball drawn at random is red.

P2=12×25×12=12×5=110

Also using the condition of a coin tossed that is ,

Step 3: The probability of drawing ball from U1, if the ball drawn at random is two white.

P3=12×3252×33=12×3!2!1!5!2!3!×1=12×3!×2!2!×3!×2×15×4×3!=12×3×25×4=35×4=320

Step 4: The probability of drawing ball from U2, if the ball drawn at random is1 red and 1 white ball.

P4=12×31×2152×23=12×3!2!1!×2!1!1!5!2!3!×23=12×3!×2!2!×2×3!×2×15×4×3!=2×25×4=45×4=15

Step 5: Also, the probability of drawing a ball from U2. If the ball drawn at random 2 is red ball

P5=12×2252×13=12×2!2!1!5!2!3!×23=12×1×3!×2!5×4×3!×13=15×4×3=160

So, required probability is 310+110+320+15+160=2330

Hence, option (B) is the correct answer


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