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**alzor****Guest**

hey guys i really need help with this q as i got an uncoming exam.

'Orchid World' nursery sells assorted orchid bulbs, 25% of which are of Phalaenopsis sort. Helena buys 8 randomly chosen bulbs without enquiring about their sort. Given that the shop assistant knows that no more than 3 of them are Phalaenopsis, the probalility that Helena bought more than 1 Phalaenopsis is...?

the answer is 0.586, but i got no idea how to get it.

plz help

thanks.

**mathsyperson****Moderator**- Registered: 2005-06-22
- Posts: 4,900

The probability of Helena choosing n Phalaenopsis orchids is 0.25^n x (1-0.25)^(8-n) x 8Cn.

There's probably an 'nCr' button on your calculator.

Anyway, substituting 0, 1, 2 and 3 into that, the probability of her choosing:

None is 0.100112915

One is 0.266967773

Two is 0.311462402

Three is 0.207641602

The question tells us that she doesn't pick any more than 3, so these are the only possibilities available.

The probabilities that we are interested in are those of picking 2 and 3. Adding these gives 0.51904004. The total probability is 0.886184692, but as we know that there are no other possibilities available, we need to scale up the total probability to 1.

This is done by multiplying by 1/0.886184692, and doing the same to the probability that we're interested in gives 0.585774059, which when rounded to a sensible accuracy is 0.586.

Why did the vector cross the road?

It wanted to be normal.

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