Thanks. Our team member toyvo deserves all the credit for that.
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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.
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However, I cannot seem to figure out how you got a few of your answers, and would love some guidance:
Sure
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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
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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.
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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
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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.
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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)
simply calculate the value of probability of getting 4 heads, 5 heads and 6 heads and add them together.
<<<vs.>>>
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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?