GMAT Math

jagdeep wrote:Can anyone answer this question please


a fair two sided coin is flipped 6 times. what is the probability that tails will be the result atleast twice but not more than 5 times

Stuart Kovinsky wrote:

GMATCHPOINT wrote:(question from someone terrible in math)

Does it make sense to apply this formula on below question from OG11? If so, how can I do it? I have tried with no success.

From OG11 page 241 ex 173

THE PROBABILITY IS 1/2 THAT A CERTAIN COIN WILL TURN UP HEADS ON ANY GIVEN TOSS. IF THE COIN IS TO BE TOSSED THREE TIMES, WHAT IS THE PROBABILITY THAT ON AT LEAST ONE OF THE TOSSES THE COIN WILL TURN UP TAILS?

A) 1/8
B) 1/2
C) 3/4
D) 7/8
E) 15/16

The formula works very well for this question. First, let's simplify.

If we want AT LEAST 1 tails, then the only thing that we DO NOT want is all 3 heads.

Remember this general formula:

Prob (want) = 1 - Prob (don't want)

Applying that to this question:

Prob (at least 1 tails) = 1 - Prob (all 3 heads)

Prob(all 3 heads) = 3C3/2^3 = 1/8

Prob (at least 1 tails) = 1 - 1/8 = 7/8

Of course, you could also answer this question without the forumla, using a bit of logic and common sense.

The only result we don't want is HHH. For 3 flips, there are 2^3 = 8 total possible outcomes. If there's 1 outcome we don't want, there must be 7 outcomes that we do want, so the chance of getting what we want is 7/8.

You could also look at the n=3 row of Pascal's Triangle. The only thing we don't want is 0 tails, so if we add up 1 tails + 2 tails + 3 tails we get 3 + 3 + 1 = 7. The whole row totals to 8, so again we end up with an answer of 7/8.

Stuart Kovinsky wrote:I just typed up this detailed explanation to address a specific question and figured that I might as well share it with everyone!

Coin flip strategies

There are numerous ways to solve coin flip questions. One of the quickest is to apply the coin flip formula.

The probability of getting exactly k results out of n flips is:

nCk/2^n

For example , if one wanted to know the probability of getting exactly 3 heads out of 4 flips:

4C3/2^4 = 4/16 = 1/4

As quick as it was to apply the formula, there's an even BETTER way to solve coin flip questions, involving memorizing a few numbers.

Here are the numbers to remember:

1 3 3 1
1 4 6 4 1
1 5 10 10 5 1

Some of you may recognize those patterns as rows of numbers from Pascal's Triangle (I swear, I came up with them first ). The Triangle has a number of uses, but for GMAT purposes one of its most useful applications is to coin flip questions.

The first row applies to 3 flip questions, the second to 4 flip questions and the third to 5 flip questions.

Let's start with the first row, 1 3 3 1, and see how it helps.

"A fair coin is flipped 3 times. What's the probability of getting exactly 2 heads?"

The entries in the row represent the different ways to get 0, 1, 2 and 3 results, respectively. In our question, we want 2 heads, so we go to the 3rd entry in the row, "3".

To find the total number of possibilities, add up the row... 1+3+3+1 = 8

So, our answer is 3/8.

Going back to our original question (exactly 3 heads out of 4 flips):

4 row is 1 4 6 4 1

For 3 heads, we use the 4th entry: 4
Sum of the row is 16
Answer: 4/16 = 1/4

Let's look at a much more complicated question:

"A fair coin is flipped 5 times. What's the probability of getting at least 2 heads?"

If we want at least 2 heads, we want 2 heads, 3 heads, 4 heads OR 5 heads. Pretty much whenever we see "OR" in probability, we add the individual probabilities.

Looking at the 5 flip row, we have 1 5 10 10 5 1. For 2H, 3H, 4H and 5H we add up the 3rd, 4th, 5th and 6th entries: 10+10+5+1=26.

Summing the whole row, we get 32.

So, the chance of getting at least 2 heads out of 5 flips is 26/32 = 13/16.

kumadil2011 wrote:How can we solve the below problem using the same formula ?

https://www.beatthegmat.com/three-fair- ... 53691.html

tx

Stuart Kovinsky wrote:
"A fair coin is flipped 3 times. What's the probability of getting exactly 2 heads?"

gmathater2 wrote:Hi Stuart, i hope you are still checking this forum. First thank you for the formula, i'm solving all the coins exercises and more! I was just wondering, using you formula is it possible to use the following for die (dice) questions nCk/6^n? if not, is there a formula to make the die problems easier?

Best Regards

gmathater2 wrote:Hi Stuart,

Sorry i'm bothering you again with the same issue.

the formula for the die problems is not really working for me. Let's take the following example:
A fair 6 sided die is rolled 4 times. what is the probability that there will be a "3" at least 3 times?

according to the formula the answer is: 4C3/6^4 + 4C4/6^4

but the correct answer is: 5 * 4C3/6^4 + 4C4/6^4 OR 4 * (1/6)^3 * (5/6) + (1/6)^4

so should we add to the formula something like (number of non chosen numbers to show)^(n-k) and the formula becomes (number of non chosen numbers to show)^(n-k) * nCk/6^n

It could be possible because for the coin it's either this or that so this added factor is always 1, but for the die there are 5 other possibilities.

cheers