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If a child flips a coin five times in a row, what is the probability

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Re: If a child flips a coin five times in a row, what is the probability [#permalink] New post   18 Aug 2019, 18:25
The only ways the child won't land a head AND a tail is if he or she lands ALL heads and ALL tails

i.e. HHHHH or TTTTT

So we can deduct the probability that these two events will occur from 1 to determine the Probability that either won't occur and thus that at least 1 head AND tail will land.

P(all heads) = (1/2)^5 * 5!/5! = 1/32
P(all tails) = (1/2)^5 * 5!/5! = 1/32
Add together= 2/32 = 1/16
deduct from 1
1-1/16
=15/16
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If a child flips a coin five times in a row, what is the probability [#permalink] New post   18 Aug 2019, 18:57
nikdiablo129 wrote:
If a child flips a coin five times in a row, what is the probability that she will receive at least one head and one tail?

A) 3/4
B) 11/12
C) 15/16
D) 31/32
E) 63/64

Guys please help me. i know only one way which is a really long method wherein you list all possibilities but given the time constraints, its not possible. please show a shorter and easy way to solve this please


No of ways to get at least 1 head and 1 tail in flip of a coin 5 times= Total no of ways to flips a coin 5 times - no of ways to flip all heads - no of ways to flips all tails= 2^5 - 1-1= 32-12= 30

Total no of ways to flip a coin 5 times = 2^5=32

Probability of flipping a coin to get at least 1 head and 1 tail = 30/32=15/16

IMO C

Alternatively

Probability of getting at least 1 head and at least 1 tail = 1 - probability of getting all heads - probability of getting all tails = 1 -1/32-1/32=1-2/3230/32=15/16

IMO C
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Re: If a child flips a coin five times in a row, what is the probability [#permalink] New post   20 Aug 2019, 00:59
Total number of outcomes = 2^5 = 32

there will be only one outcome where there result is TTTTT
and there will be only one outcome where the result is HHHHH
rest will be a combination of H and T.

Favorable outcomes 32-2 = 30

Probability = 30/32 = 15/16
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Re: If a child flips a coin five times in a row, what is the probability [#permalink] New post   01 Nov 2019, 08:12
awssc wrote:
VeritasPrepKarishma wrote:

I still don't understand why multiplication should not be used.

Probability of At Least One Head = 1 - Probability of No Heads = 1 - (1/32) = 31/32
Probability of At Least One Tail = 1 - Probability of No Tails = 1 - (1/32) = 31/32

Probability of At Least One Tail AND One Head = (31/32) * (31/32)...

I'm just confused because if I were to consider the Probability of At Least One Tail (which includes the scenario of all tails and no heads) OR the Probability of At Least One Head (which includes the scenario of all heads and no tails), then I'm not finding the probability of at least one head and one tail combination...


Same question please :

Why isn't probability of At Lease 1 head AND 1 tail = Probability of At Least 1 head x Probability of At Last 1 Tail ? i.e (31/32)*(31/32).

Bunuel GMATPrepNow chetan2u

Will appreciate a conceptual explanation and any links to relevant resources.
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If a child flips a coin five times in a row, what is the probability [#permalink] New post   01 Nov 2019, 20:46
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altairahmad wrote:
Same question please :

Why isn't probability of At Lease 1 head AND 1 tail = Probability of At Least 1 head x Probability of At Last 1 Tail ? i.e (31/32)*(31/32).

Bunuel GMATPrepNow chetan2u

Will appreciate a conceptual explanation and any links to relevant resources.



Hi,

Why it is not \(\frac{31}{32}*\frac{31}{32}\)?
The EVENT of flipping the coin is only one. At least one head and at least one tail are NOT talking of different events..

So, when will it be \(\frac{31}{32}*\frac{31}{32}\)?
If the question was that he flips coins 5 times twice. In the first he gets at least one head and in other he gets at least one tail.

What is the question here?
Quote:
If a child flips a coin five times in a row, what is the probability that she will receive at least one head and one tail?


So we are talking of SAME 5 flips where there is at least one head and at least one tail is there.

TWO ways to do it..

(I) Find when this does not happen and subtract from total ways..
When will we NOT get at least one head and at least one tail? - when all are head or all are tail. -- 2 ways
Total ways \(2*2*2*2*2 = 32\) --- since 2 outcomes for each flip
Probability that all are head or all are tail = \(\frac{2}{32}=\frac{1}{16}\), so probability that at least one head and at least one tail = \(1-\frac{1}{16}=\frac{15}{16}\)

(II) Find ways in which we have at least one head and at least one tail
1 head and 4 tails -- 5C4 = 5 ways
2 heads and 3 tails -- 5C3 = 10 ways
3 heads and 2 tails -- 5C2 = 10 ways
4 heads and 31 tail -- 5C1 = 5 ways
Total possibilities = 5+10+10+5=30
Total ways = 2*2*2*2*2=32
Probability = \(\frac{30}{32}=\frac{15}{16}\)

C
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Re: If a child flips a coin five times in a row, what is the probability [#permalink] New post   03 Nov 2019, 03:53
chetan2u wrote:
altairahmad wrote:
Same question please :

Why isn't probability of At Lease 1 head AND 1 tail = Probability of At Least 1 head x Probability of At Last 1 Tail ? i.e (31/32)*(31/32).

Bunuel GMATPrepNow chetan2u

Will appreciate a conceptual explanation and any links to relevant resources.



Hi,

Why it is not \(\frac{31}{32}*\frac{31}{32}\)?
The EVENT of flipping the coin is only one. At least one head and at least one tail are NOT talking of different events..

So, when will it be \(\frac{31}{32}*\frac{31}{32}\)?
If the question was that he flips coins 5 times twice. In the first he gets at least one head and in other he gets at least one tail.

What is the question here?
Quote:
If a child flips a coin five times in a row, what is the probability that she will receive at least one head and one tail?


So we are talking of SAME 5 flips where there is at least one head and at least one tail is there.

TWO ways to do it..

(I) Find when this does not happen and subtract from total ways..
When will we NOT get at least one head and at least one tail? - when all are head or all are tail. -- 2 ways
Total ways \(2*2*2*2*2 = 32\) --- since 2 outcomes for each flip
Probability that all are head or all are tail = \(\frac{2}{32}=\frac{1}{16}\), so probability that at least one head and at least one tail = \(1-\frac{1}{16}=\frac{15}{16}\)

(II) Find ways in which we have at least one head and at least one tail
1 head and 4 tails -- 5C4 = 5 ways
2 heads and 3 tails -- 5C3 = 10 ways
3 heads and 2 tails -- 5C2 = 10 ways
4 heads and 31 tail -- 5C1 = 5 ways
Total possibilities = 5+10+10+5=30
Total ways = 2*2*2*2*2=32
Probability = \(\frac{30}{32}=\frac{15}{16}\)

C


Doubt cleared. Thanks for the explanation.
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Re: If a child flips a coin five times in a row, what is the probability [#permalink] New post   20 Jan 2020, 08:37
VeritasKarishma

Would it be possible to use combinations here? For instance 2C1/2^5?.This would give you the probability of All heads or All tails. Can you please confirm my understanding?
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Re: If a child flips a coin five times in a row, what is the probability [#permalink] New post   20 Jan 2020, 22:07
Expert Reply
gmatapprentice wrote:
VeritasKarishma

Would it be possible to use combinations here? For instance 2C1/2^5?.This would give you the probability of All heads or All tails. Can you please confirm my understanding?


Flipping a coin 5 times involves arrangements. The number of ways in which you can get all heads is different from the number of ways in which you can get 1 Heads and 4 Tails because all heads can be arranged in only 1 way HHHHH while 1 Heads and 4 Tails can be arranged in 5 ways HTTTT, THTTT, TTHTT, TTTHT and TTTTH.
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If a child flips a coin five times in a row, what is the probability [#permalink] New post   20 Jan 2020, 23:06
Total events 2^5 = 32
there are 2 events when either all Head occur or all Tail occur i.e : HHHHH, TTTTT
else in every case either 1 or more head n tail occur
favourable cases: (32-2) =30

P= 30/32 => 15/16. Answer is C
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