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! :-)
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
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)! ]
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