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Probability 19 Online
OpenStudy (anonymous):

a box contains 25 transistors 6 of which are defective if six are selected at random find the probability that none are defective? i need help figuring out how they got 27,132/177,100

Parth (parthkohli):

1 - all 6 are defective

Parth (parthkohli):

1 - (6/25)(5/24)(4/23)(3/22)(2/21)(1/20)

OpenStudy (kropot72):

This question deals with sampling without replacement. Population size = 25 Number of successes in population = 6 Sample size = 6 Number of successes in sample (X) = 0 P(X = 0) is found from the hypergeometric distribution as follows: \[P(X=0)=\frac{\left(\begin{matrix}6 \\ 0\end{matrix}\right)\left(\begin{matrix}19 \\ 6\end{matrix}\right)}{\left(\begin{matrix}25 \\ 6\end{matrix}\right)}=\frac{\frac{19!}{6!13!}}{\frac{25!}{6!19!}}=\frac{19\times 18\times 17\times 16\times 15\times 14}{25\times 24\times 23\times 22\times 21\times 20}\]

OpenStudy (kropot72):

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