Wine is shipped from a vineyard in cases of 6 bottles. Historical data indicates

Question

Wine is shipped from a vineyard in cases of 6 bottles. Historical data indicates that, on average, 2.7% of bottles turn sour by the time they are uncorked. Calculate, to the nearest %: a) the probability that 1 case of tins wine will contain on more than 1 sour bottle. b) the probability that, if 10 cases of this wine are purchased, at least 8 of them will be 'perfect' - ie, contain 6 good bottles and 0 sour ones. c) the probability that, if 1 shipped case is opened at a time until a perfect case is found, more than 2 cases will have to be opened.

Explanation / Answer

ans=

a)
n=6
p=0.027
x <= 1
Binomial:
P(x <= 1) = P(x=0) + P(x=1)
P(x=0) = 6C0 (0.027)^0 (0.973)^6 = 0.848549
P(x=1) = 6C1 (0.027)^1 (0.973)^5 = 0.141280

Add: P( no more than 1 sour bottle) = 0.989829

b)
10 cases = 60 bottles
P( a case is perfect) = P( all perfect bottles case) = 6C0 (0.027)^0 (0.973)^6 = 0.848549
n=10
x=8,9,10
p = 0.848549

P(x=8) = 10C8 ( 0.848549)^8 (1-0.848549)^2 = 0.277442
P(x=9) = 10C9 ( 0.848549)^9 (1-0.848549)^1 = 0.345434
P(x=10) = 10C10 ( 0.848549)^10 (1-0.848549)^0 = 0.193539

Add: P( at least 8 perfect cases) = 0.816415

c)
Probability of a perfect case p = 0.848549
Let x be the number of cases required to get the first success ( all non-sour-bottles case)
P( more than 2) = 1-P( x = 1)-P(x=2)
P( x=k ) = (1-p)^(k - 1) p
P(x=1) = (1-0.848549)^0 (0.848549) = 0.848549
P(x=2) = (1-0.848549)^1 (0.848549) =0.1285136
P( more than 2) = 1- 0.848549- 0.1285136
=0.02294

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