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If six coins are flipped simultaneously, the probability ...

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by gmattesttaker2 » Sun Jun 15, 2014 10:02 pm
Hello,

Can you please tell me if my approach is correct here? I was not clear with the official explanation.

If six coins are flipped simultaneously, the probability of getting at least one heads and at least one tails is closest to:

A) 3%
B) 6%
C) 75%
D) 94%
E) 97%

OA: [spoiler]97%[/spoiler]

P( At least one heads and at least one tails ) = P (At least 1 head). P(At least 1 tail)


P( At least 1 head ) = 1 - P( No heads ) = 1 - (1/2)^6 = 1 - (1/64) = 63/64

Similarly, P ( At least 1 tail ) = 63/64

Hence, P( At least one heads and at least one tails ) = P (At least 1 head). P(At least 1 tail)
= 63/64 x 63/64
= approx. 1

Hence, 97%

I was just wondering if this is correct? Thanks for your help.

Regards,
Sri
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by GMATGuruNY » Mon Jun 16, 2014 3:26 am
gmattesttaker2 wrote: If six coins are flipped simultaneously, the probability of getting at least one heads and at least one tails is closest to:

A) 3%
B) 6%
C) 75%
D) 94%
E) 97%

OA: [spoiler]97%[/spoiler]
P(good outcome) = 1 - P(bad outcome).
Here, a GOOD outcome is to to get at least one heads and at least one tails.
Thus, a BAD outcome is to NOT get at least one heads and at least one tails -- in other words, to get the SAME result on all 6 flips (all heads or all tails).
Thus:
P(at least one heads and at least one tails) = 1 - P(all 6 flips are the same).

P(all 6 flips are the same):
The first flip can be heads or tails.
P(the 2nd flip is the same as the first) = 1/2.
P(the 3rd flip is the same as the preceding flips) = 1/2.
P(the 4th flip is the same as the preceding flips) = 1/2.
P(the 5th flip is the same as the preceding flips) = 1/2.
P(the 6th flip is the same as the preceding flips) = 1/2.
To combine these probabilities, we multiply:
1/2 * 1/2 * 1/2 * 1/2 * 1/2 = 1/32 = 3/96 ≈ 0.03.

P(at least one heads and at least one tails):
1 - 0.03 = 0.97 = 97%.

The correct answer is E.
P( At least 1 head ) = 1 - P( No heads ) = 1 - (1/2)^6 = 1 - (1/64) = 63/64

Similarly, P ( At least 1 tail ) = 63/64
There is quite a bit of overlap between the two probabilities calculated above.

P(at least 1 heads) is composed of the following outcomes:
1H, 5T
2H, 4T
3H, 3T
4H, 2T
5H, 1T
6H.

P(at least 1 tails) is composed of the following outcomes:
1T, 5H
2T, 4H
3T, 3H
4T, 2H
5T, 1H
6T.

In your solution, all of the probabilities in red are counted twice.
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If six coins are flipped simultaneously, the probability ...

by Brent@GMATPrepNow » Mon Jun 16, 2014 6:49 am
gmattesttaker2 wrote: If six coins are flipped simultaneously, the probability of getting at least one heads and at least one tails is closest to:

A) 3%
B) 6%
C) 75%
D) 94%
E) 97%
A slightly different approach (with the same answer :-))

When it comes to probability questions involving "at least," it's best to try using the complement.
That is, P(Event A happening) = 1 - P(Event A not happening)
So, here we get: P(getting at least one heads and at least one tails) = 1 - P(not getting at least one heads and at least one tails)

What does it mean to not get at least one heads and at least one tails? It means getting EITHER zero heads OR zero tails.

So, we can write: P(getting at least one heads and at least one tails) = 1 - P(getting zero heads OR zero tails)
= 1 - P(getting ALL tails OR ALL heads)

P(getting ALL tails OR ALL heads)
= (tails on 1st toss AND tails on 2nd toss AND tails on 3rd toss AND tails on 4th toss AND tails on 5th toss AND tails on 6th toss OR heads on 1st toss AND heads on 2nd toss AND heads on 3rd toss AND heads on 4th toss AND heads on 5th toss AND heads on 6th toss)
= [1/2 x 1/2 x 1/2 x 1/2 x 1/2 x 1/2] + [1/2 x 1/2 x 1/2 x 1/2 x 1/2 x 1/2]
= [1/64] + [1/64]
= 2/64
= 1/32

So, P(getting at least one heads and at least one tails) = 1 - 1/32
= 31/32
≈ [spoiler]97%[/spoiler]
= E

Cheers,
Brent
Last edited by Brent@GMATPrepNow on Mon Apr 23, 2018 12:17 pm, edited 1 time in total.
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by prachi18oct » Mon Jun 16, 2014 9:59 pm
Can I use the below approach as well??

P(Atleast 1 head and atleast 1 tail) => 1-P(All heads or All tails)
=> only when HHHHHH or TTTTTT comes then we will not get atleast one head and one tail
=>Total ways of choosing 6 objects from set of 12 = 12C6 P(HHHHHH or TTTTTT) = 924
=> Total ways of getting HHHHHH or TTTTTT =>2
=>1-2/924
However I am getting almost 99%
Please explain where I did it wrong?Is this approach not correct?

I think I have taken incorrectly the total ways of selecting from 12 objects..Is it because they are similar?12C6 will have similar selections also .How can I correct this?I have to remove the similar selections but how?
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by [email protected] » Mon Jun 16, 2014 10:15 pm
Hi prachi18oct,

Yes, you can use the approach that you described.

You have made a math error though. Since there are 6 coins and each coin will either be a "heads" or a "tails", there are NOT 12 items to choose from. The possible number of outcomes is 2^6 = 64.

You are correct that there are only 2 ways of NOT getting what you "want" (at least 1 head and at least 1 tail).

Thus, there are 2/64 of NOT getting what you want and 1 - 2/64 = 62/64 = 31/32 ways of getting what you want.

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by Anantjit » Thu Jul 28, 2016 9:51 am
Brent@GMATPrepNow wrote:
gmattesttaker2 wrote: If six coins are flipped simultaneously, the probability of getting at least one heads and at least one tails is closest to:

A) 3%
B) 6%
C) 75%
D) 94%
E) 97%
A slightly different approach (with the same answer :-))

When it comes to probability questions involving "at least," it's best to try using the complement.
That is, P(Event A happening) = 1 - P(Event A not happening)
So, here we get: P(getting at least one heads and at least one tails) = 1 - P(not getting at least one heads and at least one tails)

What does it mean to not get at least one heads and at least one tails? It means getting EITHER zero heads OR zero tails.

So, we can write: P(getting at least one heads and at least one tails) = 1 - P(getting zero heads OR zero tails)
= 1 - P(getting ALL tails OR ALL heads)

P(getting ALL tails OR ALL heads)
= (tails on 1st toss AND tails on 2nd toss AND tails on 3rd toss AND tails on 4th toss AND tails on 5th toss AND tails on 6th toss OR heads on 1st toss AND heads on 2nd toss AND heads on 3rd toss AND heads on 4th toss AND heads on 5th toss AND heads on 6th toss)
= [1/2 x 1/2 x 1/2 x 1/2 x 1/2 x 1/2] + [1/2 x 1/2 x 1/2 x 1/2 x 1/2 x 1/2]
= [1/64] + [1/64]
= 2/64
= 1/32

So, P(getting at least one heads and at least one tails) = 1 - 1/32
= 31/32
≈ [spoiler]97%[/spoiler]
= A

Cheers,
Brent

Hi Brent,

How did AND become OR in complement.

So is it everytime while we take a complement of AND CONDITION it becomes OR condition?
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by Brent@GMATPrepNow » Thu Jul 28, 2016 10:35 am
Anantjit wrote:
Hi Brent,

How did AND become OR in complement.

So is it everytime while we take a complement of AND CONDITION it becomes OR condition?
Rather than try to create rules for each situation, I think it's better to understand what the complement means in each case.

For example, if a question asks us to find P(at least one head) then we can say:
P(at least one head) = 1 - P(NOT at least one head)
= 1 - P(zero heads)
= 1 - P(all tails)

Another example: if a question asks us to find P(at least two heads) then we can say:
P(at least two heads) = 1 - P(NOT at least two heads)
= 1 - P(1 head OR 0 heads)

Does that help?

Cheers,
Brent
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by Matt@VeritasPrep » Thu Aug 04, 2016 9:09 pm
Anantjit wrote:
How did AND become OR in complement.

So is it everytime while we take a complement of AND CONDITION it becomes OR condition?
x AND y is actually a subset of x OR y, at least if we're using OR in the logical sense, where OR = "at least one of ..."

For instance, suppose I'm thinking about cars and trucks, some of which are fast and some of which are big. Cars that are fast AND big are a subset of cars that are fast OR big, since they appear in the fast set and in the big set.
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