gmatt14 wrote:

A bowl contains 10 apples, 2 of which are bad. If someone randomly selected four apples from the bowl, what is the probability that at least one of the apples is bad?

Correct Answer: 2/3

I found this approach:

First selecting the probability of all good apples, then subtracting that from one in order to find the probability of at least selecting one bad apple.

P(all good)=

(8/10)*(7/9)*(6/8)*(5/7) = 1/3

P(all good)= 1/3

P(at least one bad)= 1-P(all good)

P(at least one bad)= 1-(1/3)

P(at least one bad)= 2/3

Can anybody suggest an alternative approach?

the one stated above by u is the best approach.

Alternative approach::

atleast one means case1+case 2

case1-- one bad 9 good total ways = 2c1*8c3

case2 2 bad 8 go0d total ways = 2c2*8c2

total ways = 10c4

probability = \(\frac{{2c1*8c3 + 2c2*8c2}}{10c4}\) = 2/3

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