cg-tower.com->Probability-and-statistics-> SOLUTION: A box consists of 8 blue marbles, 4 red marbles, 3 environment-friendly marbles. A. If three marbles space selected in ~ random, what is probability the all 3 marbles will certainly be blue marblesB. I var visible_logon_form_ = false;Log in or register.Username: Password: it is registered in one simple step!.Reset your password if girlfriend forgot it."; return false; } "> log in On


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Click here to check out ALL troubles on Probability-and-statisticsQuestion 267894: A box contains 8 blue marbles, 4 red marbles, 3 eco-friendly marbles. A. If 3 marbles room selected at random, what is probability the all three marbles will be blue marblesB. If three marbles space selected at random, what is the probability that one of each shade marble will certainly be selected.C. If 11 marbles room selected at random, what is the probability the no red marbles will be selected.D. If 6 marbles room selected at random, what is the probability that 2 of each color marble will certainly be selected? answer by palanisamy(496)
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(Show Source): You can put this solution on her website! A box contains 8 blue marbles, 4 red marbles, 3 eco-friendly marbles. Total number of balls = 8+4+3 = 15A. If three marbles space selected at random,what is probability the all 3 marbles will be blue marblesTotal number of ways = 15C3 = (15*14*13) / (1*2*3) = 455Out the 5 blue balls 3 deserve to be selected in 5C3 ways.The number of favourable means = 5C3 = (5*4*3)/(1*2*3)=10The probability the all three marbles will be blue marbles = 10/455 = 2/91B. If 3 marbles room selected at random,what is probability that one of each colour be selected .Total variety of ways = 15C3 = (15*14*13) / (1*2*3) = 455Out the 8 blue balls 1 deserve to be selected in 8 ways,Out the 4 red balls 1 can be selected in 4 ways,Out that 3 green balls 1 deserve to be selected in 3 ways, The variety of favourable means = 5C3 = (8*4*3)=96The probability that all 3 marbles will certainly be blue marbles = 96/455 C.

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If 11 marbles space selected in ~ random, what is the probability the no red marbles will be selectedTotal number of ways = 15C11 = 15C4 = (15*14*13*12)/(1*2*3*4) = 1365Leaving the 4 red balls there space 8 blue and also 3 green balls.So the end of 11 (blue balls+green ) 11 deserve to be selected in 1 way.The variety of favourable ways = 1The probability the all 11 marbles will be blue marbles or green = 1/455D. If six marbles room selected in ~ random, what is the probability that 2 of each shade marble will certainly be selected? Total number of ways = 15C6 = (15*14*13*12*11*10)/(1*2*3*4*5*6) = 5005Out of 8 blue balls 2 can be selected in 8C2 ways,Out of 4 red balls 2 deserve to be selected in 4C2 ways,Out that 3 green balls 2 can be selected in 3C2 ways, The number of favourable methods = (8C2)*(4C2)*(3C2)=504 The probability that that two of each shade marble will be selected= 504/5005
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