Lesson Binomial distribution problem on M&M peanuts
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<H2>Binomial distribution problem on M&M peanuts</H2> <H3>Problem 1</H3>According to Masterfoods, the company that manufactures M&M’s, 12% of peanut M&M’s are brown, 15% are yellow, 12% are red, 23% are blue, 23% are orange and 15% are green. You randomly select six peanut M&M’s from an extra-large bag of the candies. (a) Compute the probability that exactly three of the six M&M’s are blue. (b) Compute the probability that three or four of the six M&M’s are blue. (c) Compute the probability that at most three of the six M&M’s are blue. (d) Compute the probability that at least three of the six M&M’s are blue. (e) If you repeatedly select random samples of six peanut M&M’s, on average how many do you expect to be blue? (f) With what standard deviation? <B>Solution</B> <pre> (a) In this case, we have a binomial distribution with n=6 trials, k=3 successful trials. p=0.23 probability of the individual success in each trial. Apply the standard formula for this probability P = {{{C[6]^3*0.23^3*(1-0.23)^3}}} = {{{((6*5*4)/(1*2*3))*0.23^3*0.77^3}}} = 0.1111 (rounded). <U>ANSWER</U> (b) In this case P = P(3) + P(4) = {{{C[6]^3*0.23^3*0.77^3}}} + {{{C[6]^4*0.23^4*0.77^2}}} = = {{{20*0.23^3*0.77^3 + 15*0.23^4*0.77^2}}} = 0.1360 (rounded). <U>ANSWER</U> (c) Continue in the same style P = P(0) + P(1) + P(2) + P(3) = . . . = 0.972 (rounded). <U>ANSWER</U> To replace monotonic calculations, you may use very convenient online calculator https://stattrek.com/online-calculator/binomial/ (d) Continue in the same style P = P(3) + P(4) + P(5) + P(6) = 1 - ( P(0) + P(1) + P(2) ) = 1 - 0.8609 = 0.1391 (rounded). <U>ANSWER</U> To replace monotonic calculations, you may use very convenient online calculator https://stattrek.com/online-calculator/binomial/ (e) Mathematical expectation is n*p = 6*0.23 = 1.38 blue peanuts. 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