Posted by **Anonymous** on Wednesday, June 3, 2009 at 6:40pm.

1. A die labelled A has one face marked 1, one face marked 2 and the other four faces marked 3. A die labelled B has four faces marked 2 and two faces marked 3. The dice A and B are thrown. Calculate the probability is 5.

I got 18/36

2. There are 12 boys and 20 girls in a class. 25% of boys and 25% of girls study physics; What is the probability that a student chosen at random is a girl studying physics.

I found that 3/12 boys therefore study physics and 5/20 girls therefore study physics. So 9/12 and 15/20 don't study. I'm not sure what to do next...add? multiply?

3. Calculate the probability that there are fewer boys than girls in a family of 6 children if the probability of have a boy is 1/2.

?

4. There are 6 nesting boxes in a garden. The probability of one nesting bow being occupied is 0.7%; Calculate the probability that at best 5 boxes are occupied.

Therefore 99.3% not occupied ...

Could someone please read through my work and help if possible. I know there are 4 questions its more advice/tips and is this right?

Thanks for any help! :-)

- maths -
**Damon**, Wednesday, June 3, 2009 at 7:40pm
1) Yes, agree 18/36 but that is 1/2

1/6*2/6 + 4/6*4/6 = 2/36+16/36 = 18/36 = 1/2

2) 5 girls take physics. There are 32 students so 5/32

- maths -
**Damon**, Wednesday, June 3, 2009 at 7:49pm
0, 1, 2 boys out of 6

binomial distribution

p(k) = C(n,k) (1/2)^k * (1/2)^(n-k)

p(0 boys) = C(6,0) (1/2)^0 (1/2)^6

= (1/2)^6 = 1/64

p(1 boy) = C(6,1)(1/2)^1 (1/2)^5

= 6 (1/2)(1/32)

= 6/64

p(2 boys)= 15 etc

then add those three probabilities

If you do not use Pascal's triangle, the formula for C(n,k) is:

n! / [ k! (n-k)! ]

- maths -
**Damon**, Wednesday, June 3, 2009 at 7:58pm
5) I THINK YOU MEAN 0.7, NOT .7%

Do binomial distribution again but this time for 1 out of 6 unoccupied

Probability of any box being empty = .3

Probability of exactly 1 out of 6 being empty

= C(6,1)(.3)^1 (.7)^5

= 6 * .3 * .16807

- maths -
**Damon**, Wednesday, June 3, 2009 at 8:02pm
Oh, and subtract that from one to get the probability of 1, 2, 3, 4 or 5 being occupied, but not all six.

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