If six coins are flipped simultaneously, the probability of

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.

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.

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%

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

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

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

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

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%

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:

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

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

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

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


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

Regards,
Fabio.

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

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