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# These are two probability questions from GMAT club

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These are two probability questions from GMAT club [#permalink]

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13 May 2004, 08:34
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These are two probability questions from GMAT club probability course. Can anybody explain these with concepts for a beginner? Thanks.

1. If the coin is tossed 5 times, what is the probability that at least 3 out of 5 times it will show heads? (Binominal Distribution)

2. There are 2 green, 3 red, and 2 blue balls in a box. 4 are drawn at random without replacement. What is the probability that of the 4 drawn balls two are red, 1 is green, and 1 is blue? (Hypergeometric Distribution)
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13 May 2004, 08:58
I will tr my first cut. If you cannot get it then I will give another explanation.

You can choose 3 days out of 5 in 5C3 ways. These are your desired combinations.
P = Desired combinations(events)/ total combinations(events)

if you have 2 days in which it can either rain or not you have following combinations

D1 D2(days)
N N (event)
N R
R N
R R
Basically you have X^Y combinations X = no of choices( rain, no rain)
and Y = no of days ( d1 and d2 )
for 5 days you have 2^5 = 32

----------------------------------------------------------------------------
if you cannot understand this imagine you have 2 digits and they can represent numbers from 00 to 99 [total=100 numbers ]
Each digit can have 10 choices (0,1,2...9) and there are 2 digits
so you have 10^2 combinations = 100

-----------------------------------------------------------------------------

COming back to our problem
P = 10/32

Let me know if it is not clear.
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Re: Probability concepts [#permalink]

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13 May 2004, 08:59
mirhaque wrote:
These are two probability questions from GMAT club probability course. Can anybody explain these with concepts for a beginner? Thanks.

1. If the coin is tossed 5 times, what is the probability that at least 3 out of 5 times it will show heads? (Binominal Distribution)

2. There are 2 green, 3 red, and 2 blue balls in a box. 4 are drawn at random without replacement. What is the probability that of the 4 drawn balls two are red, 1 is green, and 1 is blue? (Hypergeometric Distribution)

1. If you toss a coin 5 times, then you could get head 0, 1, 2, 3, 4, 5 times. The probabilities of these events are:

0 times: each time you get NOT head, (suppose that head shows with prob. 0<p<1), so the probability of this event = (1-p)*(1-p)*(1-p)*(1-p)*(1-p). This is so because outcomes of tossing are independent and each outcome has probability of 1-p.

1 times: you have 5 options - the coin showed head in the first tossing, int the second, int the third, ... For each option, you compute the probability again using that independence rule - p*(1-p)^4. 5 times this leads to an answer 5*p*(1-p)^4.

..... etc.

2. This problem resembles that with cards in a deck. There were 52 cards (7 balls in this case), 13 hearts and 39 non-hearts (in this case - 3 red, 2 green, and 2 blue from total 7 balls). Then you start taking without replacement: you should just remeber the formula. Or, if you know that formulas are hard to remember for you, you can use the following method:

P(r = 2, g = 1, b = 1) = P(first is green)*P(r = 2, g = 1, b = 1| first is green) + P(first is red)*P(r = 2, g = 1, b = 1| first is red) + P(first is blue)*P(r = 2, g = 1, b = 1| first is blue).

P(first is red) = |red|/|total|, similarly for green and blue.

P(r = 2, g = 1, b = 1| first is green) = P(r = 2, g = 1, b = 1| first is green, second is blue)*P(second is blue| first is green) + P(r = 2, g = 1, b = 1| first is green, second is red)*P(second is red| first is green).

and so on. So, you clearly should remeber formulas to easily(and, it's important, quickly) calculate it all...
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13 May 2004, 09:05
anandnk wrote:
I will tr my first cut. If you cannot get it then I will give another explanation.

You can choose 3 days out of 5 in 5C3 ways. These are your desired combinations.
P = Desired combinations(events)/ total combinations(events)

if you have 2 days in which it can either rain or not you have following combinations

D1 D2(days)
N N (event)
N R
R N
R R
Basically you have X^Y combinations X = no of choices( rain, no rain)
and Y = no of days ( d1 and d2 )
for 5 days you have 2^5 = 32

----------------------------------------------------------------------------
if you cannot understand this imagine you have 2 digits and they can represent numbers from 00 to 99 [total=100 numbers ]
Each digit can have 10 choices (0,1,2...9) and there are 2 digits
so you have 10^2 combinations = 100

-----------------------------------------------------------------------------

COming back to our problem
P = 10/32

Let me know if it is not clear.

I see official explanation:
for a coin flip p = 50%; 5 flips means that n = 5. The probability to find is for the coin to have heads at least 3 times. The answer is:

P = 5P3 + 5P4 + 5P5 = 10/32 + 5/32 + 1/32 = 16/32 = 1/2
5P3 = 5C3 * 50% ^ 3 * (1 - 50%) ^ (5 - 3) = 10 * 1/8 * 1/4 = 10/32
5C3 = 5! / (3! * 2!) = 10
5P4 = 5C4 * 50% ^ 4 * (1 - 50%) ^ (5 - 4) = 5 * 1/32 = 5/32
5C4 = 5! / (4! * 1!) = 5
5P5 = 50% ^ 5 = 1/32
So, the answer is 1/2.
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13 May 2004, 09:18
mirhaque wrote:
anandnk wrote:
I will tr my first cut. If you cannot get it then I will give another explanation.

You can choose 3 days out of 5 in 5C3 ways. These are your desired combinations.
P = Desired combinations(events)/ total combinations(events)

if you have 2 days in which it can either rain or not you have following combinations

D1 D2(days)
N N (event)
N R
R N
R R
Basically you have X^Y combinations X = no of choices( rain, no rain)
and Y = no of days ( d1 and d2 )
for 5 days you have 2^5 = 32

----------------------------------------------------------------------------
if you cannot understand this imagine you have 2 digits and they can represent numbers from 00 to 99 [total=100 numbers ]
Each digit can have 10 choices (0,1,2...9) and there are 2 digits
so you have 10^2 combinations = 100

-----------------------------------------------------------------------------

COming back to our problem
P = 10/32

Let me know if it is not clear.

I see official explanation:
for a coin flip p = 50%; 5 flips means that n = 5. The probability to find is for the coin to have heads at least 3 times. The answer is:

P = 5P3 + 5P4 + 5P5 = 10/32 + 5/32 + 1/32 = 16/32 = 1/2
5P3 = 5C3 * 50% ^ 3 * (1 - 50%) ^ (5 - 3) = 10 * 1/8 * 1/4 = 10/32
5C3 = 5! / (3! * 2!) = 10
5P4 = 5C4 * 50% ^ 4 * (1 - 50%) ^ (5 - 4) = 5 * 1/32 = 5/32
5C4 = 5! / (4! * 1!) = 5
5P5 = 50% ^ 5 = 1/32
So, the answer is 1/2.

mirhaque, you've just found the probability of AT LEAST 3 flips occuring, while Anand says that the probability of EXACTLY 3 flips is 10/32.

By the way, if you want to find the probability of at least 3 flips, you shouldn't do all these calculations. JUST look and see, that this prob. is 1/2, because for each outcome with at least 3 flips there IS an exacly opposite outcome with no more than 2 flips:

(flip, *, flip, *, flip) ~~~ (*, flip, *, flip, *). Why you can do so when it comes to probabilities? Because p(flip) = 1/2 and flip and non-flip have equal probabilities!

For every outcome with <= 3 flips we can find an outcome with no more than 2 flips! => the number of these outcomes ARE EQUAL. but their sum = |all outcomes|, since outcomes with <= 2 flips and outcomes with >= 3 flips constitute ALL POSSIBLE outcomes. => probabilities of these events are equal = 1/2!
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13 May 2004, 12:44
Mirhaque,

What's the answer to the second problem? 12/35?
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13 May 2004, 13:14
ndidi204 wrote:
Mirhaque,

What's the answer to the second problem? 12/35?

Official answer says: we have 3 ball colors, or 3 sub-events: g for green, r for red, b for blue. We know that:

g' = 2
r' = 3
b' = 2
g = 1
r = 2
b = 1
Now, let's do the calculations:

p = 2C1 * 3C2 * 2C1 / (2+3+2)C(1+2+1) = 2C1 * 3C2 * 2C1 / 7C4
7C4 = 7! / (4! * 3!) = 5 * 6 * 7 / 6 = 35
2C1 = 2
3C2 = 3! / (2! * 1!) = 3
p = 2 * 2 * 3 / 35 = 12/35
So, the answer is 12/35.

I didn't understand this p = 2C1 * 3C2 * 2C1 / (2+3+2)C(1+2+1) part. can anybody please explain?
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13 May 2004, 13:24
No of ways to choose 2 red from 3 red = 3C2
No of ways to choose 1 green from 2 green = 2C1
No of ways to choose 1 blue from 2 blue = 2C1
Total no of ways = 3C2 * 2C1 * 2C1 = 12

No of ways to choose 4 balls out of 7 = 7C4 = 35

p = 12/35
13 May 2004, 13:24
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# These are two probability questions from GMAT club

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