probability help....?

Question

find the probability of the event
On a hospital floor, 16 patients have a disease with a mortality rate of .1. Two of them die.
and
A 10-question multiple choice test has 4 possible answers for each question. A student selects at least 6 correct answers.

please explain how to do them not just the anwser i want to learn for future problems...

Answer 1

For the 1st question, you use the binomial distribution to get the following answer:

16C2 x .1^ 2 x .9^14

Let me try to explain this. If you want 2 people to die, then 14 must live. So this basic probability is
.1^2 times .9^14
However, there a number of different ways this can happen (
The 1st 2 can die, the 1st & 4th can die, the 5th & 10th can die, etc). The number of different ways you can pick 2 out of 16 is the symbol 16C2 (16 things taken 2 at a time).
( 16C2 = (16x15)/(1x2) = 8x15=120)
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For the 2nd question: each question has a .25 chance of being right. It's the same sort of deal. The basic probability is
.25^6 x .75^4. This has to be multiplied by 10C4 which is
(10x9x8x7)/(1x2x3x4) because there are this # of different ways of selecting 4 different questions from 10. Okay?

Answer 2

As the previous responder noted, use binomial distribution to solve. The best estimate for p (death) is 0.1, and for q ( survive)=0.9. Then each patient is a "trial", and (0.1+0.9)^16=1. This can be expanded as
0.1^(16) + 16 0.1^15 x 0.9 + (16x15/2!) 0.1^14 0.9^2+ .... + (16x15/2!) 0.1^2 0.9^14 + 16 (0.9)^15 x 0.01 + 0.9^(16).
You need to know the significance of each term.
The last term is the probability that everyone will survive. The next-to-last term is the probability that exactly 15 of 16 will survive. The term before that is the term you want, the probability that exactly 14 of 16 will survive.

The second question is slightly different. p=1/4, that someone will answer a question correctly by a random pick of answer, and q=3/4, the prob that the person picks a wrong answer. So you need to expand (1/4+3/4)^10. You need to evaluate the terms for a person selects 10 correct answers, 9 correct answer, 8 correct answers, 7 correct answers, and 6 correct answers. The sum of these 5 terms is the probability you want.

Answer 3

the first is a binomial distribution with n = 16 and p = .1

16 choose 2 * .1^2 * .9 ^ 14 where .1 is chance of dying and there are 2 of those and .9 is chance of not dying (1 - .1) and there are 14 of those(16 - 2)

doing the calculation shows that .2745 chance that 2 of the 16 die

the second is also binomial with n = 10 and p = .25(1/4)

so it's 10 choose 6 * .25^6 * .75^4

this equals .016222 so there's a little less than 2% chance that just by guessing the student answers 6 correctly