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What is the probability of tossing a coin five times and having heads

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What is the probability of tossing a coin five times and having heads  [#permalink]

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New post 17 Jun 2009, 06:10
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What is the probability of tossing a coin five times and having heads appear at least two times?

A) 1/32
B) 1/16
C) 15/16
D) 13/16
E) 7/8
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Re: What is the probability of tossing a coin five times and having heads  [#permalink]

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New post 17 Jun 2009, 06:54
If 'n' coins are tossed there will
nc0 outcomes in which n coins will show head
nc1 outcomes in which 1 head will appear
and so on

Here 5 coins are tossed so the total outcomes = 2 ^5 = 32

Outcomes in which atleast 2 heads will appear = 32 - 5c0 - 5c1 = 26

Prob atleast two heads will appear = 26/32 = 13/16
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Re: What is the probability of tossing a coin five times and having heads  [#permalink]

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New post 17 Jun 2009, 07:52
Sorry man, but I dont get this

If 'n' coins are tossed there will
nc0 outcomes in which n coins will show head
nc1 outcomes in which 1 head will appear
and so on


Would you please explain me better,

Is there any other way easier

YOu are correct with your answer but I can-t understand it yet

HELP!

hehe
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Re: What is the probability of tossing a coin five times and having heads  [#permalink]

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New post 17 Jun 2009, 20:00
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Let take an example if 2 coins are tossed the total outcome= 2 ^n = 2 ^2 = 4 ( HH,HT,TH,TT)

The number of outcome 0 head can appear = nc0 = 2c0 = 1 ( TT )
The number of outcome 1 head can appear = nc1 = 2c1 = 2 (HT,TH)
The number of outcome 2 head can appear = nc2 = 2c2 = 1 (HH)

The same rule i have generalised and put below. U need to follow the same for 5 coins.

I have proved it above for 2 coins so same for other values of n.

Hope u got it.
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Re: What is the probability of tossing a coin five times and having heads  [#permalink]

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New post 18 Jun 2009, 08:31
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Here is another way to do it, although longer, but it may help you:

Total outcomes = 2^5 = 32
Outcomes which 2 heads appear = 5C2 = 5!/2!3! = 10
Outcomes which 3 heads appear = 5C3 = 5!/3!2! = 10
Outcomes which 4 heads appear = 5C4 = 5!/4!1! = 5
Outcomes which 5 heads appear = 5C5 = 5!/5!0! = 1

Sum selected outcomes = 10+10+5+1 = 26

Selected outcomes / total outcomes = 26/32 = 13/16
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New post 18 Jun 2009, 09:33
Thank you man ,

Now it did Help

I don't know how to get better in probability it is painful for me

DO you know if Prob, combinatorics and perm show very often on the GMAT?
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Re: What is the probability of tossing a coin five times and having heads  [#permalink]

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New post 18 Jun 2009, 12:09
I haven't taken the exam yet, but in my practice exams they see to show up pretty regularly (GMATPrep) and they are in the OG books (which are older GMAT questions). If I were you I would keep working on them, it took me a while to grasp the concepts but there are a lot of great posts on here which will help you as well as external links that people have posted for additional help.
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Re: What is the probability of tossing a coin five times and having heads  [#permalink]

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New post 22 Jun 2009, 07:23
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I finally knew how to solve this onein an easier way

first of all lets get what we don-t want...

I don't want TAILS so

T T T T T = 1/32

and I don-t want one head either

H T T T T = 5/32

SO I don-t want either a 1/32 or 5/32

so we get 6/32

and now 32/32 - 6/32 = 26/32

getting final answer 13/16

Very interesting question

What do you think?
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Re: What is the probability of tossing a coin five times and having heads  [#permalink]

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New post 24 Mar 2010, 14:16
TheRob wrote:
I finally knew how to solve this onein an easier way

first of all lets get what we don-t want...

I don't want TAILS so

T T T T T = 1/32

and I don-t want one head either

H T T T T = 5/32

SO I don-t want either a 1/32 or 5/32

so we get 6/32

and now 32/32 - 6/32 = 26/32

getting final answer 13/16

Very interesting question

What do you think?


I agree. One way to look at these "at least" probability questions is to take 1 - P(event not happening)

In this case, the probability of at least two heads is 1 - P(less than two heads), which means
1 - [P(zero heads) + P(one head)] which is

= 1 - (5C0/32 + 5C1/32)
= 1 - (1/32 + 5/32)

= 26/32

13/26

answer D
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Re: What is the probability of tossing a coin five times and having heads  [#permalink]

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New post 25 Mar 2010, 02:19
In questions involving Coins, its always safe to use to combinations which are less in number...

for example in question above we need to find out probability of at least 2 heads, which also means that we can just find probability when number of heads <2 i.e. no heads at all or just 1 head...then whatever is the outcome we just need to subtract it from 1.

total outcome = 1 - P(no heads+ 1head)
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Re: What is the probability of tossing a coin five times and having heads  [#permalink]

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New post 25 Mar 2010, 14:39
tkarthi4u, unless I'm doing this wrong that method doesn't work on this example: Let's say 5 coins are tossed, what is the probablility that 3 will be tails?

Total outcome = 32

Outcome of 3 tails appearing = 32-5C0-5C1-5C2 = 16
16/32 = 1/2 but that isn't the correct answer

Please advise.
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Re: What is the probability of tossing a coin five times and having heads  [#permalink]

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New post 26 Mar 2010, 16:35
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Friends, seems like there is a lot of confusion regarding questions of this type.

This question is a typical binomial probability question. Situations where there are only two possibilities that can happen, a coin is flipped and the outcome is either a heads or tails. A baby is born, and the possibility is a boy or a girl. This can however be extended to situations when a dice is rolled. Probability of getting a particular number against probability of not getting the desired number.

Now, let's extrapolate the situation to multiple trials. Let's say n trials were made, and we define a particular occurrence as a success event, the non-occurrence is thus a failure event, and r such successes are counted. The probability of each success event is p, and therefore, the probability of each failure event is q = 1-p. We need to identify the probability of r successes in n trials.

This particular situation was studied by a famous mathematician by the name Bernoulli. Hence, these are called Bernoulli's trials. And the formula goes - [nCr]*[p^(r)]*[q^(n-r)].

When a coin is flipped 5 times, probability of 2 heads/tails, note that the success events can be random, the order of success events is not considered in this situation, would be [5C2]*[(1/2)^(2)]*[(1/2)^(3)] = 5/16.

Consider a dice is rolled 8 times, probability of getting a 4 exactly 3 times would be [8C3]*[(1/6)^(3)]*[(5/6)^(5)].

Now, occurrence of a particular event at least r times would mean, non-occurrence of the event for n-r times. We can count accordingly, depending on values for n and r. Sum of all probabilities of desired events, r ranging from 0 to n, will be equal to 1.
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Re: What is the probability of tossing a coin five times and having heads  [#permalink]

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New post 27 Mar 2010, 04:25
BarneyStinson wrote:
Friends, seems like there is a lot of confusion regarding questions of this type.

This question is a typical binomial probability question. Situations where there are only two possibilities that can happen, a coin is flipped and the outcome is either a heads or tails. A baby is born, and the possibility is a boy or a girl. This can however be extended to situations when a dice is rolled. Probability of getting a particular number against probability of not getting the desired number.

Now, let's extrapolate the situation to multiple trials. Let's say n trials were made, and we define a particular occurrence as a success event, the non-occurrence is thus a failure event, and r such successes are counted. The probability of each success event is p, and therefore, the probability of each failure event is q = 1-p. We need to identify the probability of r successes in n trials.

This particular situation was studied by a famous mathematician by the name Bernoulli. Hence, these are called Bernoulli's trials. And the formula goes - [nCr]*[p^(r)]*[q^(n-r)].

When a coin is flipped 5 times, probability of 2 heads/tails, note that the success events can be random, the order of success events is not considered in this situation, would be [5C2]*[(1/2)^(2)]*[(1/2)^(3)] = 5/16.

Consider a dice is rolled 8 times, probability of getting a 4 exactly 3 times would be [8C3]*[(1/6)^(3)]*[(5/6)^(5)].

Now, occurrence of a particular event at least r times would mean, non-occurrence of the event for n-r times. We can count accordingly, depending on values for n and r. Sum of all probabilities of desired events, r ranging from 0 to n, will be equal to 1.


Nice explanation regarding Binomial distribution.
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Re: What is the probability of tossing a coin five times and having heads  [#permalink]

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New post 12 Nov 2015, 04:17
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TheRob wrote:
What is the probability of tossing a coin five times and having heads appear at least two times?

A) 1/32
B) 1/16
C) 15/16
D) 13/16
E) 7/8


The best and the fastest way to do a problem which talks about at least scenarios is to subtract the probability of that scenario happening from the total frequency.
Note: The probability of the scenario that you have to subtract will almost always be of a small number of cases.


Coming to this question:

Total cases = 2^5 = 32

Case 1: No heads = 5C0 = 1
Case 2: 1 heads = 5C1 = 5

Hence The cases which we need are 32 - 1 - 5 = 26

Probability = 26/32 = 13/16
Option D
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Re: What is the probability of tossing a coin five times and having heads  [#permalink]

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New post 10 Jul 2017, 10:00
We need to calculate the probability of having at least 2 heads after tossing a coin 5 times.

=> Calculate the probability of having 0 head and probability of having 1 head, then deduct that from 1 to find the answer.

Probability of having 0 head (T-T-T-T-T)= [(1/2)^5] = 1/32
Probability of having 1 head = [(1/2)^5]*5(positions of head) = 5/32

=> Probability of having at least 2 heads after tossing 5 coins = 1 - 1/32 -5/32 = 26/32 = 23/32
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Re: What is the probability of tossing a coin five times and having heads   [#permalink] 10 Jul 2017, 10:00
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