# Consider 10 independent tosses of a biased coin with the probability of Heads at each toss equal to p, where 0<p<1.

#### williamrobertsuk

##### New member
Consider 10 independent tosses of a biased coin with the probability of Heads at each toss equal to p, where 0<p<1.
1.Let A be the event that there are 6 Heads in the first 8 tosses. Let B be the event that the 9th toss results in Heads.
=p

2.Find the probability that there are 3 Heads in the first 4 tosses and 2 Heads in the last 3 tosses. Express your answer in terms of p using standard notation. Remember not to use ! or combinations in your answer.
=12*p^5*(1-p)^2

3.Given that there were 4 Heads in the first 7 tosses, find the probability that the 2nd Heads occurred at the 4th toss. Give a numerical answer.
=9/35

4.We are interested in calculating the probability that there are 5 Heads in the first 6 tosses and 3 Heads in the last 5 tosses. Give the exact numerical values of a, b, c, d that would match the answer ap7(1−p)3+bpc(1−p)d.
a=
b=
c=
d=

I have managed to find the answers for the questions 1, 2, 3 but number 4 tricks me a bit, someone that can help? I attached a screenshot of question 4 too. Thanks!

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#### Romsek

##### Full Member
4 is pretty straightforward. The tosses are independent so you can consider the first 6 and the last 5 as totally separate from one another.

I suspect you know how to calculate the probability of 5 heads out of 6 and 3 heads out of 5. Do this and multiply them.

you end up with an expression that looks like

$$\displaystyle \dbinom{n_1}{k_1}\dbinom{n_2}{k_2}p^{k_1+k_2}(1-p)^{(n_1-k_1)+(n_2-k_2)}$$

and I leave it to you to figure out those constants and match them to your letters.

#### pka

##### Elite Member
Consider 10 independent tosses of a biased coin with the probability of Heads at each toss equal to p, where 0<p<1.
2.Find the probability that there are 3 Heads in the first 4 tosses and 2 Heads in the last 3 tosses. Express your answer in terms of p using standard notation. Remember not to use ! or combinations in your answer.
To williamrobertsuk: One comment & one question.
Comment. If a coin the tossed or flipped it is sent spinning in the air and caught in the fly. Under those conditions a group of engineers in CA proved it is impossible to have a biased coin.

Question: Where are you( school, state or country) where standard notation means "not to use ! or combinations in your answer"?

#### Romsek

##### Full Member
Question: Where are you( school, state or country) where standard notation means "not to use ! or combinations in your answer"?
That probably just means the problem writer wants these expanded out into numbers.

#### pka

##### Elite Member
That probably just means the problem writer wants these expanded out into numbers.
Given the fact that there is a WolfFramAlpha app for smart watches, testing companies are moving to more symbolic answers. Actually preliminary result has shown a better result at testing actual content knowledge.

#### williamrobertsuk

##### New member
4 is pretty straightforward. The tosses are independent so you can consider the first 6 and the last 5 as totally separate from one another.

I suspect you know how to calculate the probability of 5 heads out of 6 and 3 heads out of 5. Do this and multiply them.

you end up with an expression that looks like

$$\displaystyle \dbinom{n_1}{k_1}\dbinom{n_2}{k_2}p^{k_1+k_2}(1-p)^{(n_1-k_1)+(n_2-k_2)}$$

and I leave it to you to figure out those constants and match them to your letters.
I will give it a shot! Thanks

#### williamrobertsuk

##### New member
To williamrobertsuk: One comment & one question.
Comment. If a coin the tossed or flipped it is sent spinning in the air and caught in the fly. Under those conditions a group of engineers in CA proved it is impossible to have a biased coin.

Question: Where are you( school, state or country) where standard notation means "not to use ! or combinations in your answer"?
Standard notation means to use combinations!

#### williamrobertsuk

##### New member
I am actually struggling a bit to find out!

#### williamrobertsuk

##### New member
I managed to calculate C=8! Any help with the rest?!

#### Dr.Peterson

##### Elite Member
I managed to calculate C=8! Any help with the rest?!
4.We are interested in calculating the probability that there are 5 Heads in the first 6 tosses and 3 Heads in the last 5 tosses. Give the exact numerical values of a, b, c, d that would match the answer a p7(1−p)3+bpc(1−p)d.
a=
b=
c=
d=
I think you're referring to the value of "c" (not "C") here.

Note that I have put exponents in your expression where I assume they belong; that is important.

What's more important than the numerical answers is your thinking. Whatever thinking you did to decide that c = 8 might be important either in determining whether you are thinking in the right direction, or in helping you move on from there.

So please tell us what you understand about the problem, and what you have done to try to solve it.

You should probably not be looking at the requested form, but just trying to solve the problem itself, and then match your answer up with the form. In particular, as you have already been told, you can start by telling us how you'd find the probability of getting 5 heads in 6 tosses of this hypothetical coin.

But also please check that all the numbers in the problem as you gave it are correct. Something doesn't make sense.

#### pka

##### Elite Member
I managed to calculate C=8! Any help with the rest?!
4.We are interested in calculating the probability that there are 5 Heads in the first 6 tosses and 3 Heads in the last 5 tosses. Give the exact numerical values of a, b, c, d that would match the answer ap7(1−p)3+bpc(1−p)d.
a=
b=
c=
d=
$$\displaystyle \left\{ {\binom{6}{5}{p^5}{{(1 - p)}^1}} \right\}\left\{ \binom{5}{3}p^3(1-p)^2 \right\}=60p^8(1-p)^{3}$$

#### williamrobertsuk

##### New member
What confuses me is the fact that the two events overlap, that makes it tricky for me!
d=2

Last edited:

#### pka

##### Elite Member
What confuses me is the fact that the two events overlap, that makes it tricky for me!/QUOTE]
4.We are interested in calculating the probability that there are 5 Heads in the first 6 tosses and 3 Heads in the last 5 tosses. Give the exact numerical values of a, b, c, d that would match the answer ap7(1−p)3+bpc(1−p)d.
I see absolutely no possible overlapping. There a string of six followed by a string of five. The two strings are disjoint.

#### Dr.Peterson

##### Elite Member
Consider 10 independent tosses of a biased coin with the probability of Heads at each toss equal to p, where 0<p<1.
...
4.We are interested in calculating the probability that there are 5 Heads in the first 6 tosses and 3 Heads in the last 5 tosses. Give the exact numerical values of a, b, c, d that would match the answer ap7(1−p)3+bpc(1−p)d.
What confuses me is the fact that the two events overlap, that makes it tricky for me!
Ah! I missed that there are only 10 tosses in all, so that the 6th toss affects both parts. You're right.

What you might do is to split the compound event into two cases (heads on toss 6, tails on toss 6) and add them together.

Give that a try, and show some work.

#### pka

##### Elite Member
Ah! I missed that there are only 10 tosses in all, so that the 6th toss affects both parts. You're right.
What you might do is to split the compound event into two cases (heads on toss 6, tails on toss 6) and add them together.
Give that a try, and show some work.
Consider 10 independent tosses of a biased coin with the probability of Heads at each toss equal to p, where 0<p<1.
1.Let A be the event that there are 6 Heads in the first 8 tosses. Let B be the event that the 9th toss results in Heads.
=p
2.Find the probability that there are 3 Heads in the first 4 tosses and 2 Heads in the last 3 tosses. Express your answer in terms of p using standard notation. Remember not to use ! or combinations in your answer.
=12*p^5*(1-p)^2

3.Given that there were 4 Heads in the first 7 tosses, find the probability that the 2nd Heads occurred at the 4th toss. Give a numerical answer.
=9/35

4.We are interested in calculating the probability that there are 5 Heads in the first 6 tosses and 3 Heads in the last 5 tosses. Give the exact numerical values of a, b, c, d that would match the answer ap7(1−p)3+bpc(1−p)d.
a=
b=
c=
d=
I have managed to find the answers for the questions 1, 2, 3 but number 4 tricks me a bit, someone that can help? I attached a screenshot of question 4 too.
If one looks at the attachment, the screenshot, in the OP it seems clear that #4 is not part if the rest. Or at least that is how I read it.