If six coins are flipped simultaneously, the probability of

Question

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%

MacFauz · Accepted Answer

Probability of getting no heads = 1/64
Probability of getting at least 1 head = 63/64

Probability of getting no tails = 1/64
Probability of getting at least 1 tail = 63/64

Probability of getting at least 1 head and at least 1 tail = 63*63/64*64
63/64 is greater than 98%... So squaring that should be greater than 96%. Answer is E.

BrentGMATPrepNow · 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)
=  x  1/2  x  1/2  x  1/2  x  1/2  x  1/2]   +    x  1/2  x  1/2  x  1/2  x  1/2  x  1/2] 
=     +    
= 2/64
=  1/32

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

Answer: E

Cheers,
Brent

BabySmurf · Answer

Probability of getting one head (or one tail) = 1/2.

Case one:  1st die is head. Probability of at least one of the remaining five to be tail = 1 - P(none tail).
Probability(none tail i.e. all head) = 1/(2^5) = 1/32. Thus, P(at least one tail) = 31/32.
Probability of  case one  that one is head and at least another is tail, 
i.e. at least one head and one tail = 1/2 * 31/32 = 31/64.

Case two:  1st die is tail. Probability of at least one of the remaining five to be head = 1 - P(none head).
Probability(none head i.e. all tail) = 1/(2^5) = 1/32. Thus, P(at least one head) = 31/32.
Probability of  case two  that one is tail and at least another is head, 
i.e. at least one tail and one head = 1/2 * 31/32 = 31/64. .....we don't need to do this second case calculation
actually since we know it will be the same result as tail and head have same probability.

Since the events that could happen are Case One  or  Case Two we have to ADD these two probabilities:
31/64 + 31/64 = 62/64.

Two question regarding this:
1. Is there a  faster  approach than the aforementioned (assuming the rationale is correct).

2. How do I figure out  quickly  that 62/64 = 31/32 = 0.96875 = 97%?

I was lucky on this practice question as I was running out of time, and my gut feeling said it has to be close to 99%.
But just rounding down the numerator to make division easier ... 30/32 = 0.9375 = 94% answer would change to D.

Correct answer is btw. E.

farful · Answer

1. Yes. You can reword the question to: What is the probability you won't get all heads or all tails?

Should be fairly straightforward that probability of all heads is 1/(2^6). Likewise for all tails. so 1/64 + 1/64 = 2/64... and thus the probability of not getting all heads or all teails is 1 - 1/32 or 31/32.

2. 32 is close to 33... and 33*3 is pretty close to 100. So I would do 31/32 = 93/96. Since we're just four off, 93+4 / 96+4 or 97/100 should be close enough.

m3equals333 · Answer

Yup, I attempted to use the reverse combinatorics approach (I think that's what it is called) and it was pretty quick.

total potential outcomes (H or T @6 Coins): 2^6 = 64

outcomes that don't include at least one heads and one tails:

\HHHHHH
TTTTTT

so two. then take 1-P(not HT)

1 - 2/64 = 31/32 = 93/96 = or very close to 97%

christianbze · Answer

Instead of assuming that  \frac{31}{32}  is 97%, you could do the following operation:

\frac{31}{32} = 1 - \frac{1}{32} ;

100 : 32 = 3,1...  -->  1 : 32 = 0,031...

shasadou · Answer

hi all

can i use the  combinations with repeating elements  approach here?

clearly we meet the condition only if get either H H H H H H or T T T T T T out of a large number of possible combinations. If you have 30 secs left, play the game - choose between D and E.

1. As long as we have 6 elemens that can repeat: we have 6 * 6 * 6 * 6 * 6 * 6 total number of combinations.
2. HHHHHH in the combinatorics language means  \frac{6!}{1!}  = 6! = 720. TTTTTT is the same: 720. Sum up the 2: 1440
3. find the fraction via factorization:  \frac{12*12*10}{3^6*2^6}  =  \frac{5}{162}  which is obviosly is less than 6%=0.06
4. E is correct

zxcvbnmas · Answer

Total outcomes= 2*2*2*2*2*2 = 64
these outcomes wont work(where all heads or all tails)= 2
so not getting the outcome=  \frac{2}{64} 
getting the outcome =  1 - \frac {2}{64}  -->  \frac{31}{32}

\frac{31}{32}  = 0.96 =97%

EMPOWERgmatRichC · Answer

Hi All,

When dealing with probability questions, there are only two results that can be calculated: what you WANT to have happen OR what you DON'T WANT to have happen. Those two outcomes create the following equation:

(Want) + (Don't Want) = 1

Sometimes it's actually easier to calculate what you WANT to have happen by calculating what you DON'T WANT to have happen (and then subtract that fraction from the number 1).

Here, we're asked for the probability of flipping AT LEAST one head and AT LEAST one tail on 6 coin flips. Since each coin flip has 2 possible outcomes, there are 2^6 = 64 possible outcomes (although there would be lots of 'duplicate results'). We don't want to have to determine every possible outcome that gives us at least 1 head and at least 1 tail though, so let's calculate the probability of that NOT happening.

There are two results that would NOT fit what we're looking for:

ALL HEADS 
ALL TAILS

The probability of each is the same: 1/64

1/64 + 1/64 = 2/64 = 1/32

1/32 = about 3%

(Want) + (.03) = 1

Want = about 97%

Final Answer:  E

GMAT assassins aren't born, they're made, 
Rich

fskilnik · Answer

?\,\,\, = \,\,\,1 - P\left( {{	ext{in}}\,\,6\,\,{	ext{flips}},\,\,6H\,\,{	ext{or}}\,\,6T} ight)\,\,\, = \,\,\,1 - {?_{{	ext{temporary}}}}

\left. \begin{gathered}
 {	ext{Total}} = \,\,{2^6}\,\,\,{	ext{equiprobables}} \hfill \
 {	ext{Favorable}} = \,2\,\,\,\,\left( {6H\,\,{	ext{or}}\,\,6T} ight)\,\, \hfill \ 
\end{gathered} ight\}\,\,\,\, \Rightarrow \,\,\,\,{?_{{	ext{temporary}}}} = \frac{2}{{{2^6}}} = \frac{1}{{32}} = \frac{{96 + 4}}{{32}}\% = 3\frac{1}{8}\%

?\,\,\, \cong \,\,\,100\% - 3\%

This solution follows the notations and rationale taught in the GMATH method.

Regards, 
Fabio.

ScottTargetTestPrep · Answer

There are 2^6 = 64 ways the six coins could be flipped. Of the these 64 ways, only two of them (all heads and all tails) do not have at least one head or at least one tail. Therefore, the probability of getting at least one head and at least one tail is 62/64 = 0.96875 ≈ 97%.

Answer: E

BrushMyQuant · Answer

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

6 coins are tossed => Total number of cases =  2^6  = 64

Out of the these 64 cases we have TTTTTT, HHHHHH, and 62 cases where we have at least 1 Tail and at least 1 Head

=> P(At least 1T and At least 1H) =  \frac{62}{64}  =  \frac{31}{32}  ~ 97%

So,  Answer will be E 
Hope it helps!

Watch the following video to learn How to Solve Probability with Coin Toss Problems

https://www.youtube.com/watch?v=LRURv38kZw8

bumpbot · Answer

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If six coins are flipped simultaneously, the probability of

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