GMAT Math

Hi mariolita,

If you can post the question here (either take a picture or transcribe it), then I'm sure you'll get some responses.

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

mariolita wrote: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

I believe this is the question you're referring to:

The probability is ½ 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

Here's one approach:

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 1 tails) = 1 - P(not getting at least 1 tails)
What does it mean to not get at least 1 tails? It means getting zero tails.
So, we can write: P(getting at least 1 tails) = 1 - P(getting zero tails)

Now let's calculate P(getting zero tails)
What needs to happen in order to get zero tails?
Well, we need heads on the first toss and heads on the second toss and heads on the third toss.
We can write P(getting zero tails) = P(heads on 1st AND heads on 2nd AND heads on 3rd)
This means that P(getting zero tails) = P(heads on 1st) x P(heads on 2nd) x P(heads on 3rd)
Which means P(getting zero tails) = (1/2)x(1/2)x(1/2)= 1/8

We're now ready to answer the question.
P(getting at least 1 tails) = 1 - P(not getting at least 1 tails)
= 1 - 1/8
= [spoiler]7/8[/spoiler]
= D

Cheers,
Brent

The probability is ½ 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

Another approach is to draw a TREE DIAGRAM that shows all of the possible outcomes.

Here's the 1st toss:


Then add the 2nd toss:


Then the 3rd:


So, as you can see there are 8 possible outcomes:



Now let's place a checkmark beside those outcomes that satisfy the condition of having AT LEAST one tails.


As we can see, 7 of the 8 outcomes satisfy this condition. So, the probability is [spoiler]7/8[/spoiler]

Answer: D

Cheers,
Brent

Hi ocina,

This is a layered probability question that requires some additional math (the combination formula, or the equivalent "mapping" of all possibilities).

We're told that the coin will be flipped 5 times; with every "win", Kate gets $1 and with every "loss", Kate loses $1. Kate starts with $10. We're asked for the probability that Kate ends up with MORE than $10 but less than $15 after 5 tosses.

Let's start with the total number of possible outcomes. Since each coin has 2 options (heads or tails), there are 2^5 = 32 possible outcomes for the 5 flips (which will include a certain number of similar outcomes in different orders - for example HHTTT and THTHT).

To end up with MORE than $10, Kate has to win MORE tosses than she loses. However, if she were to win all 5 tosses, she'd have $15 and we want her to end up with LESS than $15. This means that Kate has to win EITHER 3 times or 4 times.

Since it does not matter which of the 5 tosses is won, as long as it's either 3 or 4 of them, we can use the combination formula:

Combinations = N!/[K!(N-K)!]

For 3 wins, we have 5!/3!2! = 10 possible combinations of 3 wins

For 4 wins, we have 5!/4!1! = 5 possible combinations of 3 wins

So, there are a total of 10+5 = 15 combinations of 5 tosses that "fit" what we're looking for and 32 possible outcomes total.

Final Answer: [spoiler]15/32[/spoiler]

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