Statistics simple probability of tossing a coin

Leah.J

13

If a fair coin is tossed 10 times,then the probability of having at most 5 heads is ?

How do I solve this ?

(edited 4 years ago)

Reply 1

Notnek

21

Original post by Leah.J

If a fair coin is tossed 10 times,then the probability of having at most 5 heads is ?

How do I solve this ?

Are you familiar with the binomial distribution? If not, what level of maths are you doing?

Reply 2

Leah.J

OP

13

Original post by Sir Cumference

Are you familiar with the binomial distribution? If not, what level of maths are you doing?

I finished a level maths but I don't remember the binomial distribution.

I feel like it has sth to do with permutations and combinations but I don't know how to use these to get a probability. I only took S1, we didn't take permutations and combinations.

Original post by Leah.J

If a fair coin is tossed 10 times,then the probability of having at most 5 heads is ?

How do I solve this ?

You have a chance of 1/2 to get heads on a single toss.

If you want to get heads 0 times, then this event is the same as getting all tails (T) {T,T,T,T,T,T,T,T,T,T}. This is the only case in which you get 0 heads. What is the probability of getting this??

Now, if you want 1 head, then this event is the same as getting {H,T,T,T,T,T,T,T,T,T}. What is the probability of this case happening?? By extension, notice that this scenario says that heads must come from the fist toss, but obviously the reality is that heads can come up on any one of these 10 drops. It come up as a 2nd position {T,H,T,T,T,T,T,T,T,T} ... or 3rd position {T,T,H,T,T,T,T,T,T,T}, etc... So, in terms of the choose function, how many arrangements can we have of this event?? Hence, what is the probability of this event happening overall??

Repeat similarly for 2 heads, 3 heads, 4 heads, and 5 heads. Add up all of these probabilities.

Reply 4

Notnek

21

Original post by Leah.J

I finished a level maths but I don't remember the binomial distribution.

I feel like it has sth to do with permutations and combinations but I don't know how to use these to get a probability. I only took S1, we didn't take permutations and combinations.

So is this a question from a university course then? Isn't the binomial distribution part of that course? You could try it without knowing the binomial distribution but you'll end up learning the binomial distribution in the process :smile:

So I'd recommend starting by learning/understanding the binomial distribution and finding out how to calculate e.g. P(getting exactly 5 heads). Then you'll need to add these up

P(0 heads) + P(1 head) + P(2 heads) + P(3 heads) + P(4 heads) + P(5 heads)

Reply 5

Leah.J

OP

13

Original post by RDKGames

You have a chance of 1/2 to get heads on a single toss.

If you want to get heads 0 times, then this event is the same as getting all tails (T) {T,T,T,T,T,T,T,T,T,T}. This is the only case in which you get 0 heads. What is the probability of getting this??

Now, if you want 1 head, then this event is the same as getting {H,T,T,T,T,T,T,T,T,T}. What is the probability of this case happening?? By extension, notice that this scenario says that heads must come from the fist toss, but obviously the reality is that heads can come up on any one of these 10 drops. It come up as a 2nd position {T,H,T,T,T,T,T,T,T,T} ... or 3rd position {T,T,H,T,T,T,T,T,T,T}, etc... So, in terms of the choose function, how many arrangements can we have of this event?? Hence, what is the probability of this event happening overall??

Repeat similarly for 2 heads, 3 heads, 4 heads, and 5 heads. Add up all of these probabilities.

I get what you're saying and I analyzed it that way but I didn't know whether to use nCr or nPr, why did you choose nCr
and is it gonna be
1/2^10 +10C1/2^10 +10C2/2^10 ... 10C5/2^10 ?

Original post by Leah.J

I get what you're saying and I analyzed it that way but I didn't know whether to use nCr or nPr, why did you choose nCr
and is it gonna be
1/2^10 +10C1/2^10 +10C2/2^10 ... 10C5/2^10 ?

Yes.

Well you use nCr because when it comes to figuring out the number of ways H appears once in a list of 10 slots, you're essentially asking yourself "in how many ways can I choose one spot out of 10?" and just put H in that spot.

Similarly for when you're dealing with 2 heads. You're asking "in how many ways can I choose two spots out of 10?" and just put H in them. Order does not matter therefore you do not need to use nPr.

And so on...

(edited 4 years ago)

Reply 7

Leah.J

OP

13

Original post by RDKGames

Yes.

Well you use nCr because when it comes to figuring out the number of ways H appears once in a list of 10 slots, you're essentially asking yourself "in how many ways can I choose one spot out of 10?" and just put H in that spot.

Similarly for when you're dealing with 2 heads. You're asking "in how many ways can I choose two spots out of 10?" and just put H in them. Order does not matter therefore you do not need to use nPr.

And so on...