Bowl one (B1) has 2 blue chips and 4 red chips, bowl two (B2) has 3 blue chips a

Question

Bowl one (B1) has 2 blue chips and 4 red chips, bowl two (B2) has 3 blue chips and 2 red chips, and bowl three (B3) has 4 blue chips and 1 red chip. P(B1)=3/5, P(B2)=1/10, and P(B3)=3/10. Find the probability of bowl one is chosen given that a red chip is drawn. Bowl one (B1) has 2 blue chips and 4 red chips, bowl two (B2) has 3 blue chips and 2 red chips, and bowl three (B3) has 4 blue chips and 1 red chip. P(B1)=3/5, P(B2)=1/10, and P(B3)=3/10. Find the probability of bowl one is chosen given that a red chip is drawn.

Explanation / Answer

P(bowl 1 | red chip) = P(bowl 1 and red chip)/P(red chip)

= (3/5 x 4/6)/{(3/5 x 4/6) + (1/10 x 2/5) + (3/10 x 1/5)}

= 0.4/(0.4 + 0.04 + 0.06)

= 0.8

Probability of bowl one is chosen given that a red chip is drawn = 0.8

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