A bag contains 3 red balls, 2 white balls and 3 blue balls. Three balls are selected randomly from the bag with replacement. Given that no blue ball has been selected, calculate the probability that the number of red balls selected exceeds the number of white balls.

ans choices: (a) 3/8 (b) 3/5 (c) 7/10 (d) 81/512 (e) 81/125

My solution: Want P(R>W|B'), where R = # of red, W = # of white, B = # of blue = P(R>W and B')/ P(B')

= [(3/8)^2(2/8)*3!/2!1! + (3/8)^3] / (1-(3/8)^3)

= not the answer given above.

reasoning: (3/8)^2 * 2/8 multiplied by 3!/2! since this is the probability of getting two red and 1 white, added with 3/8^3 since this is the probability of choosing three red.

The denominator is due to the fact that to pick no blue is to find 1-prob of choosing all three blue

Thank you

ans choices: (a) 3/8 (b) 3/5 (c) 7/10 (d) 81/512 (e) 81/125

My solution: Want P(R>W|B'), where R = # of red, W = # of white, B = # of blue = P(R>W and B')/ P(B')

= [(3/8)^2(2/8)*3!/2!1! + (3/8)^3] / (1-(3/8)^3)

= not the answer given above.

reasoning: (3/8)^2 * 2/8 multiplied by 3!/2! since this is the probability of getting two red and 1 white, added with 3/8^3 since this is the probability of choosing three red.

The denominator is due to the fact that to pick no blue is to find 1-prob of choosing all three blue

Thank you

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