'A box contains 12 bulbs, 4 of them are defective. We select 3 bulbs randomly from the box and X denotes the number of defective bulbs among the three selected bulbs. The probability P(X-3) is equal to'
Added by -Scar S.
Step 1
This can be done using the combination formula: C(12,3) = 12! / (3! * 9!) = 220 So, there are 220 ways to select 3 bulbs from 12 bulbs. Show more…
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