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Two candy bowls contain red and green candies. The first bowl contains an even mix of red and green​ candies, while the second bowl contains 30​% red and 70​% green candies. A bowl is​ chosen, and a candy from that bowl is randomly selected. Suppose that whenever a red candy is​ chosen, it came from the first bowl 75.0​% of the time. What is the probability that the first bowl is​ chosen? g

Respuesta :

Answer: Our required probability is 0.642.

Step-by-step explanation:

Since we have given that

Probability of red candies from first bowl = 50% = 0.5

Probability of green candies from second bowl = 50% = 0.5

Let the probability of chosing first bowl be 'x'.

Let the probability of chosing second bowl be '1-x'

Probability of red candies from second bowl = 30% = 0.3

Probability of green candies from second bowl = 70% =0.7

Probability of red that comes from the first bowl = 75% = [tex]P(B_1|R)=0.75[/tex]

According to question,

[tex]P(B_1|R)=\dfrac{P(R|B_1).P(B_1)}{P(R|B_1).P(B_1)+P(R|B_2).P(B_2)}\\\\0.75=\dfrac{0.5x}{0.5x+0.3(1-x)}\\\\0.75=\dfrac{0.5x}{0.5x+0.3-0.3x}\\\\0.75=\dfrac{0.5x}{0.2x+0.3}\\\\0.75(0.2x+0.3)=0.5x\\\\0.15x+0.225=0.5x\\\\0.225=0.5x-0.15x\\\\0.225=0.35x\\\\\dfrac{0.225}{0.35}=x\\\\0.642=x[/tex]

Hence, our required probability is 0.642.

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