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Probability Help!!!

by bsharad_99 » Tue May 26, 2015 5:19 pm
Hi,

It's been couple of weeks that I have started practicing for GMAT and like most of the aspiring candidates I have started with Math. Within this two week I have done fair amount of math problem and can I say I am making sound progress. One math topic I am struggling with is "Probability" and Its getting frustrating as I am not able to unlock these types of problem and derive at the correct answer. I know at least one question of Probability will show up in the exam and I don't want to lose out on scoring Probability section. I would appreciate if you can provide techniques/steps/tricks to break down Probability question.

I have done multiple questions and struggled with all of the the example question below.
Josh and Dan have 12 apples each. Together they flip a coin 6 times. For every heads, josh receives an apple from Dan and foe every tail Dan recieves an apple from Josh. After the coin has been flipped 6 times, what is the probability that josh has more than 12 apples but less than 18 apples?

A: 5/6
B: 21/64
C: 21/32
D: 1/64
E: 15/64

Thanks,
Sharad
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by GMATGuruNY » Tue May 26, 2015 7:56 pm
bsharad_99 wrote: I have done multiple questions and struggled with all of the the example question below.
Josh and Dan have 12 apples each. Together they flip a coin 6 times. For every heads, josh receives an apple from Dan and for every tail Dan recieves an apple from Josh. After the coin has been flipped 6 times, what is the probability that josh has more than 12 apples but less than 18 apples?

A: 5/6
B: 21/64
C: 21/32
D: 1/64
E: 15/64
Josh begins with 12 apples.
For every heads, Josh gets 1 apple.
For every tails, Josh loses 1 apple.
A favorable outcome occurs when Josh finishes with between 12 and 18 apples.

6 heads, 0 tails --> Josh gets 6 apples and loses 0 apples --> 12+6-0 = 18 apples.
5 heads, 1 tails --> Josh gets 5 apples and loses 1 apple --> 12+5-1 = 16 apples.
4 heads, 2 tails --> Josh gets 4 apples and loses 2 apples --> 12+4-2 = 14 apples.
3 heads, 3 tails --> Josh gets 3 apples and loses 3 apples --> 12+3-3 = 12 apples.

Only the options in red yield a favorable outcome.
Question stem, rephrased:
If a coin is flipped 6 times, what is the probability of getting 4 or 5 heads?

P = (total good outcomes)/(total possible outcomes).

Total possible outcomes:
Since there are 2 possible outcomes for each of the 6 flips, the total number of ways to flip a coin 6 times = 2*2*2*2*2*2 = 64.

Good outcomes:
From 6 flips, the number of combinations of 4 that can yield 4 heads = 6C4 = (6*5*4*3)/(4*3*2*1) = 15.
From 6 flips, the number of combinations of 5 that can yield 5 heads = 6C5 = (6*5*4*3*2)/(5*4*3*2*1) = 6.
Total good outcomes = 15+6 = 21.

Thus:
P(4 or 5 heads) = (total good outcomes)/(total possible outcomes) = 21/64.

The correct answer is B.
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by nikhilgmat31 » Tue May 26, 2015 9:20 pm
Why not probability for 5 & 4 heads is calculated as
probability of single head = 1/2
probability of not head i.e. = 1/2

(1/2)^5 *(1/2)^1 + (1/2)^4 * (1/2)^2 = 1/64 + 1/64 = 1/32


Please help me here.
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by GMATGuruNY » Wed May 27, 2015 3:12 am
nikhilgmat31 wrote:Why not probability for 5 & 4 heads is calculated as
probability of single head = 1/2
probability of not head i.e. = 1/2

(1/2)^5 *(1/2)^1 + (1/2)^4 * (1/2)^2 = 1/64 + 1/64 = 1/32


Please help me here.
The calculation in red accounts for the probability that the FIRST 5 TOSSES ARE HEADS and that the LAST TOSS IS TAILS (HHHHHT).
The calculation in blue accounts for the probability that the FIRST 4 TOSSES ARE HEADS and that the LAST TWO TOSSES ARE TAILS (HHHHTT).
But these are not the ONLY ways a favorable outcome can be yielded.
For example:
Exactly 5 heads will be yielded if the FIRST TOSS IS TAILS and the LAST 5 TOSSES ARE HEADS (THHHHH).
Exactly 4 heads will be yielded if the SECOND AND THIRD TOSSES ARE TAILS and the REMAINING TOSSES ARE HEADS (HTTHHH).
To account for ALL OF THE WAYS to get a favorable outcome, we could proceed as follows:

P(exactly n times) = P(one way) * total possible ways.

Case 1: Exactly 5 heads
P(one way):
In 6 tosses, one way to get exactly 5 heads is HHHHHT.
P(HHHHHT) = (1/2)(1/2)(1/2)(1/2)(1/2)(1/2) = 1/32.

Total possible ways:
Any arrangement of the letters HHHHHT represents one way to get exactly 5 H"s and 1 T.
Thus, to account for ALL OF THE WAYS to get exactly 5 H's and 1 T, the result above must be multiplied by the number of ways to arrange the letters HHHHHT.
Number of ways to arrange 6 elements = 6!.
But when an arrangement includes IDENTICAL elements, we must divide by the number of ways each set of identical elements can be ARRANGED.
The reason:
When the identical elements swap positions, the arrangement doesn't change.
Here, we must divide by 5! to account for the five identical H's:
6!/5! = 6.

Multiplying the results above, we get:
P(exactly 5 H's) = 6 * 1/32 = 6/32.

Case 2: Exactly 4 heads
P(one way):
in 6 tosses, one way to get exactly 4 heads is HHHHTT.
P(HHHHTT) = (1/2)(1/2)(1/2)(1/2)(1/2)(1/2) = 1/32.

Total possible ways:
Any arrangement of the letters HHHHTT represents one way to get exactly 4 H"s and 2 T's.
Thus, to account for ALL OF THE WAYS to get exactly 4 H's and 2 T's, the result above must be multiplied by the number of ways to arrange the letters HHHHTT.
Number of ways to arrange 6 elements = 6!.
But when an arrangement includes IDENTICAL elements, we must divide by the number of ways each set of identical elements can be ARRANGED.
The reason:
When the identical elements swap positions, the arrangement doesn't change.
Here, we must divide by 4! to account for the four identical H's and by 2! to account for the two identical T's:
6!/(4!2!) = 15.

Multiplying the results above, we get:
P(exactly 4 H's) = 15 * 1/32 = 15/32.

Thus:
P(exactly 5 heads or exactly 4 heads) = 6/32 + 15/32 = 21/32.

More practice:
https://www.beatthegmat.com/select-exact ... 88786.html
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https://www.beatthegmat.com/at-a-blind-t ... 20058.html
https://www.beatthegmat.com/rain-check-t79099.html
https://www.beatthegmat.com/probability-t227448.html
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by nikhilgmat31 » Thu May 28, 2015 5:24 am
Thanks for good explanation.

BUT

How does it matter that 1st TAIL comes of 5 HEADs come. Since the number of apples given/taken will remain same corresponding to each HEAD/TAIL which will be >12 & <18 ?

Totally confused.
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by GMATGuruNY » Thu May 28, 2015 6:43 am
nikhilgmat31 wrote:Thanks for good explanation.

BUT

How does it matter that 1st TAIL comes of 5 HEADs come. Since the number of apples given/taken will remain same corresponding to each HEAD/TAIL which will be >12 & <18 ?

Totally confused.
Josh will finish with 16 apples if the 6 flips yield exactly 5 heads and 1 tails.
Here are all of the ways to get exactly 5 heads and 1 tails:
HHHHHT
HHHHTH
HHHTHH
HHTHHH
HTHHHH
THHHHH.
Total ways = 6.

Since there are 6 WAYS that Josh will finish with 16 apples, and there are a total of 32 WAYS to flip a coin 6 times, we get:
P(Josh finishes with 16 apples) = 6/32.
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by nikhilgmat31 » Thu May 28, 2015 6:45 am
Thanks it helps & justifies the answer.
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