# Math Help - Questions Help

1. ## Questions Help

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 $\frac{1}{5}$. So the probability that 11 or 12 come will be
$P(not enough hats)=(\frac{4}{5})^{11}(\frac{1}{5})+(\frac{4}{5} )^{12}=0.0859$ 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 $(n\geq2)$, he scores at least two hits.
For (a), there would be the probability of hitting 4 or 5 times. $p^4(1-p)+p^5=p^4$ 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 $1-np(1-p)^{n-1}-(1-p)^n$.

For all of these where have i got wrong?
Thanks for any help!

2. Originally Posted by arze
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 $\frac{1}{5}$. So the probability that 11 or 12 come will be
$P(not enough hats)=(\frac{4}{5})^{11}(\frac{1}{5})+(\frac{4}{5} )^{12}=0.0859$
The probability that 11 or 12 come is:

$P(\text{not enough hats})=12 \times 0.8^{11}\times 0.2+0.8^{12}\approx 0.275$

CB

3. Originally Posted by arze
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 $\frac{1}{5}$. So the probability that 11 or 12 come will be
$P(not enough hats)=(\frac{4}{5})^{11}(\frac{1}{5})+(\frac{4}{5} )^{12}=0.0859$ 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.
Or,

there is a $\frac{4}{5}$ probability of a child coming and a $\frac{10}{12}$ probability of a hat being available and a $\frac{9}{10}$ probabilty of deciding to wear it.

The probability of a hat being worn by a child is $\frac{4}{5}\frac{10}{12}\frac{9}{10}$

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 $\frac{12}{10}\frac{4}{5}\frac{10}{12}\frac{9}{10}= \frac{36}{50}=\frac{72}{100}=0.72$

Therefore the probability a hat is not available to a child is 0.28

4. Originally Posted by Archie Meade
Or,

there is a $\frac{4}{5}$ probability of a child coming and a $\frac{10}{12}$ probability of a hat being available and a $\frac{9}{10}$ probabilty of deciding to wear it.

The probability of a hat being worn by a child is $\frac{4}{5}\frac{10}{12}\frac{9}{10}$

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 $\frac{12}{10}\frac{4}{5}\frac{10}{12}\frac{9}{10}= \frac{36}{50}=\frac{72}{100}=0.72$

Therefore the probability a hat is not available to a child is 0.28
There are two parts and two answers to this question. The first part does not include the probability that a child refuses to wear the hat, and the answer is 0.28. when the probability that a child refuses to wear a hat is considered, the answer is 0.11.
Thanks

5. Originally Posted by Archie Meade
Or,

there is a $\frac{4}{5}$ probability of a child coming and a $\frac{10}{12}$ probability of a hat being available and a $\frac{9}{10}$ probabilty of deciding to wear it.

The probability of a hat being worn by a child is $\frac{4}{5}\frac{10}{12}\frac{9}{10}$

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 $\frac{12}{10}\frac{4}{5}\frac{10}{12}\frac{9}{10}= \frac{36}{50}=\frac{72}{100}=0.72$

Therefore the probability a hat is not available to a child is 0.28
Pardon?!

CB

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 $\frac{1}{5}\ (\frac{4}{5})^{12}$

Since any of 12 children can choose not to arrive

$P(child\ missing)=\frac{12}{5}\ (\frac{4}{5})^{11}$

If all 12 come there will be a shortage,
this can happen in one way only.

Hence, $P(shortage)=\frac{12}{5}\ (\frac{4}{5})^{11}+(\frac{4}{5})^{12}=0.275$

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.

$P(shortage)=\frac{12}{5}(\frac{4}{5})^{11}(0.9)^{1 1}+(\frac{4}{5})^{12}[12(0.9)^{11}(0.1)+(0.9)^{12}]=0.105$

The above was edited as a factor was inadvertently omitted.

8. Originally Posted by arze

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 $(n\geq2)$, he scores at least two hits.
For (a), there would be the probability of hitting 4 or 5 times. $p^4(1-p)+p^5=p^4$ 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 $1-np(1-p)^{n-1}-(1-p)^n$.

For all of these where have i got wrong?
Thanks for any help!
The only thing you are missing is "counting the number of ways the event can happen".

Q2 (a) There are $\binom{5}{4}$ ways to miss once..

on the 1st, 2nd, 3rd, 4th or 5th shot.

$P(miss\ once)=\binom{5}{4}p^4(1-p)=5p^4(1-p)=5p^4-5p^5$

$P(at\ least\ one\ hit)=1-(5p^4-5p^5+p^5)=1-5p^4+4p^5$

Q2 (b) There are $\binom{n}{n-1}$ ways to miss n-1 times, therefore to hit once.

$P(one\ hit)=\binom{n}{n-1}p(1-p)^{n-1}=np(1-p)^{n-1}$

$P(no\ hit)=(1-p)^n$

$P(at\ least\ 2\ hits)=1-(np[1-p]^{n-1}+[1-p]^n)$

9. Originally Posted by Archie Meade
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.

$P(shortage)=\frac{12}{5}(\frac{4}{5})^{11}+(\frac{ 4}{5})^{12}[(0.9)^{11}+(0.9)^{12}]=0.105$
Can't get the last calculation you did. does not look correct either.

10. 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.

11. I've tried that equation you gave above and it still does not get the answer 0.11

12. Sorry arze,

I dropped the $0.9^{11}$ factor of $\frac{12}{5}(\frac{4}{5})^{11}0.9^{11}$

apologies again..

13. Thank you very much for all the help!

14. Hi arze,
i actually omitted 2 factors..

$P(shortfall)=\binom{12}{11}(0.2)^1(\frac{4}{5})^{1 1}(0.9)^{11}+(\frac{4}{5})^{12}(0.9)^{12}+(\frac{4 }{5})^{12}[\binom{12}{11}(0.9)^{11}(0.1)^1]$

$P(shortfall)=\frac{12}{5}(\frac{4}{5})^{11}(0.9)^{ 11}+(\frac{4}{5})^{12}[(0.9)^{12}+(1.2)(0.9)^{11}]=0.10998$