GMAT Math

Stuart Kovinsky wrote:
If we want fewer (as a sentence correction aside, we use "fewer" when the noun is countable - less vs fewer is commonly tested on the GMAT ) than 2 boys, we want 0 boys or 1 boy.

So, we go to the n=10 row (which is the 11th row down - remember that the apex of the triangle is the n=0 row) and add up the first 2 entries. The sum of the row will be 2^10.

It would actually be much quicker to do this using combinations, since we're looking at k values of 0 and 1:

10C0 = 1 (any nC0 = 1)
10C1 = 10 (any nC1 = n)

In fact, by the above principles (or just by looking at the triangle), we can see that the first two entries in every row are 1 and n (and, since the rows are symmetrical, the last two entries in every row are n and 1).

So, the answer would be:

1 - (1 + 10)/2^10

1 - 11/2^10

1024/1024 - 11/1024

(1024 - 11)/1024 = 1013/1024

amitdgr wrote:Is there a possibility of GMAT asking probability questions such as

There are 30% chances of boy being born and 70% chances of girl being born...Ten mothers are going to give birth to ten babies this morning. What is the probability that at least two babies are boys?

That would definitely complicate things a bit, Cuz we cant use the above discussed methods then.....

I hope such problems are not what GMAT is out to test us upon...

Thanks for being such a great help Stuart..

Stuart Kovinsky wrote:
Even your original question (with the 50/50 chance) is likely more difficult than anything you'll see on the GMAT. This version is WAY harder than any GMAT probability question (mostly because it requires calculations that are too time consuming to do by hand).

Ian Stewart wrote: I'm about 75% sure that I have seen a GMAT question like this. If I recall correctly, I saw a question testing this idea in GMATFocus, but I see a lot of GMAT-style questions from various sources, so I'm not completely sure that was the source (I might instead be remembering something I saw on a forum or from another place). As Stuart points out, almost any question like this would require an awkward amount of calculation to complete, but the question I recall seeing had answers that were not simplified, so it was easy to answer in under two minutes- as soon as you saw how to set up the calculation, you were done. The numbers were a bit easier than those above- the question was something more like:

In Country X, each time a family has a child, there is a 30% chance that the child is a boy and a 70% chance the child is a girl. If a family in Country X will have five children, what is the probability exactly one of the children will be a girl?

with answer choices like:

0.7
(0.3)^4
(0.7)*(0.3)^4
etc.

None of those answers is right, but I'll leave the question unsolved for now in case anyone wants to give it a shot.

amitdgr wrote:I am back with another coin-flip question. :mrgreen:

Kate and David each have $10. Together they flip a coin 5 times. Every time the coin lands on heads, Kate gives David $1. Every time the coin lands on tails, David gives Kate $1. After the coin is flipped 5 times, what is the probability that Kate has more than $10 but less than $15.

a) 5/16
b) 15/32
c) 1/2
d) 21/32
e) 11/16

mberkowitz wrote:"1 - (1 + 10)/2^10

1 - 11/2^10

1024/1024 - 11/1024

(1024 - 11)/1024 = 1013/1024"

can you explain this last step, specifically why it is 1 minus ... i would expect it to be 11/2^10 rather than 1- 11...

very helpful post thus far...

thanks.

pre-gmat wrote:Stuart,

Is the above strategy applicable to only scenarios where the probability of Heads and tails is 50% each? One of the questions i got on GMAT Practice test GMat prep 2 ( i know the answer is posted elsewhere on the forum but wanted to ask you about it) is following:

For one toss of a certain coin, the probability that the outcome is heads is 0.6. If this coin is tossed 5 times, which of the following is the probability that the outcome will be heads 'at least' 4 times?

(A) (0.6)^5

(B) 2(0.6)^4

(C) 3[(0.6)^4](0.4)

(D) 4[((0.6)^4)(0.4)] + (0.6)^5

(E) 5[((0.6)^4)(0.4)] + (0.6)^5

When I saw the coin flip question, i immediately tried using the pascals traingle. It did not work. is there a way to use the pascals strategy on this one?