I've got two questions:
1. A mother knows that 20% of children who accept invitation to birthday parties do not come. If she invites 12 children to a party and only has 10 party hats, what is the probability that there is not a hat for every child who comes?
The mother knows that there is a probability of 0.1 that a child who comes refuses to wear a hat. if this is taken into account, what is the probability that the number of hats will not be adequate?
I know that the probability of a child not coming is . So the probability that 11 or 12 come will be
the answer for this part is supposed to be 0.28, without the first part i don't see that the second part can be done.
2.When a boy fires a rifle at a range the probability that he hits the target is p.
(a) Find the probability that, firing 5 shots, he scores at least 4 hits.
(b) Find the probability that, firing n shots , he scores at least two hits.
For (a), there would be the probability of hitting 4 or 5 times. which is not the answer. Answer is [tex]5p^4-4p^5[\math]
(b) answer would be 1 minus the probability of missing all and hitting 1.
[tex]1-p(1-p)^{n-1}-(1-p)^n[\math]. answer is .
For all of these where have i got wrong?
Thanks for any help!
Or,
there is a probability of a child coming and a probability of a hat being available and a probabilty of deciding to wear it.
The probability of a hat being worn by a child is
This means the "statistical" fractional number of hats out of 10 being worn is 12 times that (it's just a way of thinking of it! as we can't have a fraction of a hat).
As there are 10 hats available, divide by 10 for the probability..
The probability of all 10 hats being worn is
Therefore the probability a hat is not available to a child is 0.28
Sincere apologies lads!
I saw 0.28 and didn't even think it through.
Red card offense, a spell on the sidelines would do me good...
If 11 or 12 children come, then there will be a shortage of hats.
Any of 12 children can decide not to come.
The probability of only one particular child not coming is
Since any of 12 children can choose not to arrive
If all 12 come there will be a shortage,
this can happen in one way only.
Hence,
Part (ii)
There is a 90% chance of a child deciding to choose a hat on arriving.
Hence, if 11 come and all decide to choose,
or, if 12 come and 11 or 12 choose a hat,
there will be a shortage.
The above was edited as a factor was inadvertently omitted.
Hi arze,
in this case, if 12 children came and 2 of them decided not to wear a hat, then the number of hats will be adequate, since if any of the other children want one, there will be enough.
There will be a shortfall only if 12 come and 12 or 11 want to wear a hat.
If 11 come, there will be a shortfall only if they all want a hat.
If 10 or less come, there is never a shortfall.
The calculations reflect 11 children arriving and all choosing to wear a hat given that the probability of wanting one is 0.9.
12 children arriving for which there is 2 cases...
(a) all 12 want to wear a hat
(b) 11 of the 12 want to wear a hat.
Only in those 3 cases would a shortfall occur.