Someone inadvertently mixed a bag of peanut M&Ms and peanut butter M&Ms in a can

Question

Someone inadvertently mixed a bag of peanut M&Ms and peanut butter M&Ms in a candy dish. Knowing that the resulting mixture is 70% peanut M&Ms what is The probability of selecting 4 M&Ms and getting 2 peanut and 2 peanut butter? Please show the work Someone inadvertently mixed a bag of peanut M&Ms and peanut butter M&Ms in a candy dish. Knowing that the resulting mixture is 70% peanut M&Ms what is The probability of selecting 4 M&Ms and getting 2 peanut and 2 peanut butter? Please show the work

Explanation / Answer

here probability of peanut =0.7

and  probability of peanut butter =1-0.7 =0.3

hence from multinomial distribution probability of selecting 4 M&Ms and getting 2 peanut and 2 peanut butter

=(4!/(2!*2!))*(0.7)2*(0.3)2 =0.2646

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