BTGmoderatorDC 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 E

Source: Veritas Prep

Instead of calculating the probability of getting at least one head and at least one tail, let's calculate the probability of getting NO head and the probability of getting NO tail. We'll then subtract these values from 1 to get the answer.

Probability of getting NO head = Probability of getting ALL tail = (1/2)^6

Similarly,

Probability of getting NO tail = Probability of getting ALL head = (1/2)^6

Thus, the probability of getting at least one head and at least one tail = 1 - (1/2)^6 - (1/2)^6 = 1 - (1/2)^5 = 1 - 1/32 = 31/32 = ~97%

The correct answer: E

Hope this helps!

-Jay
_________________
Manhattan Review GRE Prep

Locations: [url=hhttps://www.manhattanreview.id/gre-prep-jakarta/]GRE Classes Jarkarta[/url] | GRE Prep Course Kuala Lumpur | GRE Prep Manila | GRE Prep Course Bangkok | and many more...

Schedule your free consultation with an experienced GMAT Prep Advisor! Click here.

BTGmoderatorDC 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 E

Source: Veritas Prep

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
≈ 97%

Answer: E

Cheers,
Brent

BTGmoderatorDC 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%

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

Hi All,

We're told that six coins are flipped simultaneously. We're asked which probability is closest to getting AT LEAST one heads and AT LEAST one tails. This question comes down to a bit of probability math (which is essentially just about doing the necessary Arithmetic).

With 6 coins, there are (2)(2)(2)(2)(2)(2) = 64 possible outcomes. The ONLY outcomes that would not include at least one head AND at least one tail would be 'all heads' or 'all tails' --> HHHHHH or TTTTTT. Those are 2 of the 64 possible outcomes, meaning that 2/64 = 1/32 outcomes don't fit what we're looking for.

32 divides into 100 a little more than 3 times, so 1/32 = about .03 = about 3%. Thus, the other 97% of outcomes gives us what we're looking for.

Final Answer: E

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