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# Praetorian - your guides on Probabilty and Combinations are

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Manager
Joined: 20 Oct 2003
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17 May 2004, 11:45
1
KUDOS
This topic is locked. If you want to discuss this question please re-post it in the respective forum.

Praetorian - your guides on Probabilty and Combinations are just great. I cannot imagine being tested on a question that isnt rooted in the material you review in your tutorials (although I've been surprised before by this test).

However, I cannot seem to figure out how you got a few of your answers, and would love some guidance:

1) There are 5 married couples and a group of three is to be formed out of them; how many arrangements are there if a husband and wife may not be in the same group? (Ans. 80)

2) How many different signals can be transmitted by hoisting 3 red, 4 yellow and 2 blue flags on a pole, assuming that in transmitting a signal all nine flags are to be used? (Ans. 1260) - (not sure exactly what you're looking for in this question.....)

3) The probability that it will rain in NYC on any given day in July is 30%. what is the probability that it will rain on exactly 3 days from July 5 to July 10 ?

The second question I have is in regards to probability and Binomial Distribution. For example:

1) What is the probability, when flipping a coin 6 times, of getting at least 4 heads?
<<<vs.>>>
2) What is the probability, when flipping a coin 6 times, of getting exactly 4 heads?

Thanks. Again, kudos to the best probability/combinations guide on the market, IMO.
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Re: Practice Questions and Confusion on Probability Concept. [#permalink]

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18 May 2004, 02:26
Thanks. Our team member toyvo deserves all the credit for that.

Quote:
Praetorian - your guides on Probabilty and Combinations are just great. I cannot imagine being tested on a question that isnt rooted in the material you review in your tutorials (although I've been surprised before by this test).

the surprise part is because most of the times, we try to avoid solving tough problems. the truth is that the tough problems teach us a lot about concepts. In our prep, many of us do the mistake of solving by Process of Elimination or by guessing. Sure, know these "tricks", but learn the concepts thoroughly. GMAT tests your ability to perform under pressure , which is by the way, an essential trait for any professional.

Quote:
However, I cannot seem to figure out how you got a few of your answers, and would love some guidance:

Sure

Quote:
1) There are 5 married couples and a group of three is to be formed out of them; how many arrangements are there if a husband and wife may not be in the same group? (Ans. 80)

you can form a group of 3 from 10 people (5 couples =10 people) in 10C3 ways. ok?

Now, We simply calculate in how many ways a couple CAN be there. once we have that , we can subtract it from 10C3 and that would be your answer.

so, number of ways in which a couple is selected in the group of 3 = 5C1 * 8C1 = 40

( we first select a couple out of five couples, then out of the remaining 8 people, we select one to complete the groups of 3)

so, our final solution is 10C3 - 40 = 80 ways

Quote:
2) How many different signals can be transmitted by hoisting 3 red, 4 yellow and 2 blue flags on a pole, assuming that in transmitting a signal all nine flags are to be used? (Ans. 1260) - (not sure exactly what you're looking for in this question.....)

It is simply 9! / 3! *4! * 2!

There are three distinct groups that have to arranged. notice that within each group, all the flags are same. So, for example, for the 3 red flags, there are 3! = 6 ways of arranging them. But all these 6 arrangements are effectively the same as there is no difference in the signal. That is why we have to divide the 9! by 3!...Similarly,do it for red and yellow.

Quote:
3) The probability that it will rain in NYC on any given day in July is 30%. what is the probability that it will rain on exactly 3 days from July 5 to July 10 ?

there are 5C3 =10 ways in which it can rain 3 days out of 5

for each of these ways, the probability of rain on exactly 3 days is

A = 0.3 * 0.3* 0.3 * 0.7 * 0.7

Note that the probability is the same for rain whether it rains on July 5 or July 7. So for 10 ways, simply multiply the above by 10, because value of A will not change.

the answer = 10 * (0.3)^3 * (0.7)^2

Quote:
The second question I have is in regards to probability and Binomial Distribution. For example:

you dont need to know about binomial distribution. if the problem is clear, the answer becomes obvious.

Quote:
1) What is the probability, when flipping a coin 6 times, of getting at least 4 heads?

Atleast 4 heads = P(4) + P(5) + P(6)

<<<vs.>>>

Quote:
2) What is the probability, when flipping a coin 6 times, of getting exactly 4 heads?

This is like the NYC problem, would you like to try it?
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18 May 2004, 10:47
Quote:
probability of rain in July.

I understand that the probability will always be 30% in july regardless. So isn't it that the probability to rain in exactly in 3 days should be

0.3 *0.3 * 0.3 * (10) I didn't understand why we have to multiply 0.70 for 2 times. Can you explain further ?
Manager
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19 May 2004, 11:05
Great Explanations. Glad I dont have to apply the binomial distribution formula (although I've got it down, just in case...)

2) What is the probability, when flipping a coin 6 times, of getting exactly 4 heads?

= 6C4 * (1/2)(1/2)(1/2)(1/2)(1/2)(1/2)
= 15 * 1/64
= 15/64

Correct?
CEO
Joined: 15 Aug 2003
Posts: 3454
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21 May 2004, 10:21
wing wrote:
Quote:
probability of rain in July.

I understand that the probability will always be 30% in july regardless. So isn't it that the probability to rain in exactly in 3 days should be

0.3 *0.3 * 0.3 * (10) I didn't understand why we have to multiply 0.70 for 2 times. Can you explain further ?

i honestly dont know how to explain this. i hate to use the binomial explanation here, but i cant think of any other way

can anyone help wing? sorry about that

sincerely
praet
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21 May 2004, 10:44
Wing's question is

Why is the probability of rain for a particular choice is denoted as

0.3*0.3*0.3 *0.7 *0.7

ie. why should we multiply by 0.7 also. That is because the question asks for the choices where EXACTLY THREE DAYS IT RAINS ie. IT DOES NOT RAIN the remaining two days.

________________________________________
Well, let us take a simpler problem so that we are able to enumerate easily.

I have three coins. All the three coins are biased so that Probability of heads =0.3 and probability of tails =0.7

What is the probability that in three throws (with all three coins being thrown together) there are EXACTLY TWO HEADS ?

Solution : Enumerate

A throw where there are EXACTLY TWO Heads can be like
HHT
HTH
THH

Now Prob(HHT) = 0.3*0.3*0.7 = 0.063
Prob (HTH) = 0.3*0.7*0.3 = 0.063
Prob (THH) = 0.7 *0.3 *0.3 = 0.063

Total probability of TWO HEADS = 3 * 0.063 = 0.189

Is it clear why in the above example we need to multiply by 0.7 for every show of tales ? If so, the same is true for rains example.

transpose the rains problem to coins problem. you have 5 coins. All five are biased. prob(heads)=0.3 , prob (tails) = 0.7. what is the probability of EXACTLY THREE heads in 10 throws ??

= 0.3 * 0.3 * 0.3 * 0.7 * 0.7 * 10

Ok ?
Wing - Post if this is still not clear.
H
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22 May 2004, 09:19
praet & anupag

thank you so much. I understand clearly now. But how can I improve my probability & combiantion/permutation section ? Combination problems are rare on gmat but probability problems are frquent on gmat.
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22 May 2004, 10:42
Probability is a subject which requires a liitle bit more practice since there are a few concepts - ALL of which are required for solving problems. I would suggest that you please go back to 12th standard textbooks of your country. There will be Probability at that standard. Study from basic books. It may take you two hours per day for a week. DO NOT try to solve problems on this forum or GMAT questions without this practice. Also please go through the OG basic maths pages after you are through with this drill from textbooks. Then you can take Kaplan and OG questions. I am sure you will do well. There are many pages on the net also.
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22 May 2004, 15:51
Praetorian wrote:
wing wrote:
Quote:
probability of rain in July.

I understand that the probability will always be 30% in july regardless. So isn't it that the probability to rain in exactly in 3 days should be

0.3 *0.3 * 0.3 * (10) I didn't understand why we have to multiply 0.70 for 2 times. Can you explain further ?

i honestly dont know how to explain this. i hate to use the binomial explanation here, but i cant think of any other way

can anyone help wing? sorry about that

sincerely
praet

Between 5th and 10th of July, there's a total of 6 days, is there not? If that's the case, then the answer to the question should be --> 20 *.3^3*.7^3

any thoughts?
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23 May 2004, 02:43
yes , you are right, it should be six days, and accordingly the solution given above needs to be modified.
CEO
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23 May 2004, 04:59
lastochka wrote:
Praetorian wrote:
wing wrote:
Quote:
probability of rain in July.

I understand that the probability will always be 30% in july regardless. So isn't it that the probability to rain in exactly in 3 days should be

0.3 *0.3 * 0.3 * (10) I didn't understand why we have to multiply 0.70 for 2 times. Can you explain further ?

i honestly dont know how to explain this. i hate to use the binomial explanation here, but i cant think of any other way

can anyone help wing? sorry about that

sincerely
praet

Between 5th and 10th of July, there's a total of 6 days, is there not? If that's the case, then the answer to the question should be --> 20 *.3^3*.7^3

any thoughts?

ah, yes. justin and wing, please note.

the number of combinations are 6C3 = 20 , not 10.
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25 May 2004, 06:35
Guys,

Concerning Q3, lastochka gave us the right answer, which, in fact, is derived from Binomial Distribution (a.k.a. Bernulli theorem).

Here's my humble explanation of Binomial Distribution:

I'll try to translate this theorem from my native language into English.

If the probability p of occuring the event A in every experiment is CONSTANT, then the probability Pm,n that the event A will occur m times in n independent experiment is given by:

Pm,n=nCm x p^m x q^(n-m), where q=1-p.

In our case prob p is 0.3 (30%) and it's CONSTANT on any given day in July (from the stem), prob q is 0.7, n is 6 (total number of days), and m is 3 (...on exactly 3 days).

Plug these numbers in the formula given above:

Pm,n=6C3 x (0.3)^3 x (0.7)^3

My regards
25 May 2004, 06:35
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