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A competition to catch the most rainbow trout on Lake Rainbow pitched the men against the ladies. 16 men and 12 women participated. Each participant had a total of 15 attempts before they retired from the competition. It was predicted that the men would have a 9% chance of catching a rainbow trout at each attempt. One participant's experience resulted in 3 successful catches. The probability that this participant was a woman is 40%. Find the probability that a participant successfully catches 3 rainbow trout, given that the participant was a woman.

Sagot :

The probability that a participant successfully catches 3 rainbow trout, given that the participant was a woman is 0.0042.

P(woman) = 12/28

= 3/7

P(men) = 16/28

= 4/7

P(3 trouts) = P(3 trout | women)*P(woman) + P(3 trout | men)*P(man)

= P(3 trout | women)*(3/7) + (15C3 * (0.09^3)*(1-0.09)^(15*3))*(4/7)

= P(3 trout | women)*(3/7) + (0.00476)*(4/7)

P(woman | 3 trout) = P(3 trout | women)*P(woman) / P(3 trouts)

{P(3 trout | women) = p}

p*(3/7) / (p*(3/7) + (0.00476)*(4/7)) = 0.40

p*(3/7)/0.40 = (p*(3/7) + (0.00476)*(4/7))

p*((3/7)/0.40 - (3/7)) = (0.00476)*(4/7)

p = [(0.00476)*(4/7)] / [((3/7)/0.40 - (3/7))]

= 0.0042

Learn more about probability here https://brainly.com/question/24756209

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