There are two boxes: Box A and Box B. Box A contains 5 statistic books and 3 calculus books. Box B contains 4 statistic books and 2 calculus books. A coin is thrown twice. If at least 1 Head appears, a book from box A will be taken. Otherwise, a book from box B will be taken. a. If 2 books are taken, what is the probability that those are calculus books?

chrisysakh

chrisysakh

Open question

2022-09-03

There are two boxes: Box A and Box B. Box A contains 5 statistic books and 3 calculus books. Box B contains 4 statistic books and 2 calculus books. A coin is thrown twice. If at least 1 Head appears, a book from box A will be taken. Otherwise, a book from box B will be taken. a. If 2 books are taken, what is the probability that those are calculus books?

Answer & Explanation

geoiste72

geoiste72

Beginner2022-09-04Added 4 answers

I don't think your answer to a) is correct. The problem is quite complicated, but basically the coin tossing results in 3/4 probability of taking a book from box A and a 1/4 probablity of taking a book from box B.When it says "two books" are taken, I assume they mean that the coin tossing happens twice, so you have:(3/4)(3/4) = 9/16 that both books come from box A2(3/4)(1/4) = 6/16 that one book comes from box A and one from box B((1/4)(1/4) = 1/16 that both books come from box BI would blacko your calculations based on this.Also, although you get the right answer using binomials, this seems to me a complicated way to do it. The probability of picking two calculus books from box A (given that both books come from box A) is simply:(3/8)(2/7) = 6/56 = 3/28And two calculus books from box B (given both books come from box B) is simply:(2/6)(1/5) = 2/20 = 1/15And one calculus book from A and one from B (given that one book comes from each box) is:(3/8)(2/6) = 1/8You can put this altogether now to get the probability of two calculus books being chosen as:(Note: this is where I used the conditional probailities calculated above:)(9/16)(3/28) + (1/16)(1/15) + (6/16)(1/8)

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