If a bowl contains ten hazelnuts and eight almonds, what is the probability that four nut randomly selected from the bowl will all be hazelnuts?

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Answer 1 · 2018-03-28T07:49:45

$0.06862745 ~~6.9%$

We can model the bowl of nuts with a hypergeometric distribution containing 10 "successes" (hazelnuts) and 8 "failures" (almonds).

The formula we want to use is

$P(x=m)=(C_m^M*C_(n-m)^(N-M))/(C_n^N)$

Where:

$N=18$ (the population size)
$M=10$ (the number of successes in the population)

$n=4$ (the sample size)
$m=4$ (the number of successes we want in the sample)

Then the probability of getting 4 hazelnuts in a random sample of 4 nuts is:

$P(x=4)=(C_4^10*C_0^8)/(C_4^18)=(210*1)/(3060)~~0.06862745~~6.9%$