# Statistics simple probability of tossing a coin

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

How do I solve this ?

How do I solve this ?

Last edited by Leah.J; 1 year ago

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#2

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

How do I solve this ?

**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 ?

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(Original post by

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

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

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.

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#4

**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 ?

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.

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#5

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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.

**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.

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

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(Original post by

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.

**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.

and is it gonna be

1/2^10 +10C1/2^10 +10C2/2^10 ... 10C5/2^10 ?

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#7

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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 ?

**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 ?

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...

Last edited by RDKGames; 1 year ago

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(Original post by

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...

**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...

How many 3 digit even numbers can we form using numbers 1,2,3,4,5 ?

50 ? I did 5*5*2

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#9

(Original post by

I wanna make sure of sth:

How many 3 digit even numbers can we form using numbers 1,2,3,4,5 ?

50 ? I did 5*5*2

**Leah.J**)I wanna make sure of sth:

How many 3 digit even numbers can we form using numbers 1,2,3,4,5 ?

50 ? I did 5*5*2

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