Problem Solving

ash g wrote:Hey Stuart,

Regarding
"(Further, we predict that the answer should be a bit more than 2/3 - (e) really seems too big, so (d) looks like the best guess.) "


I think
[1] these probability problems would normally be in 700-800 range and to be good at strategic guessing to deal with such problems would be a very good skill to have - quickkill.
[2] this usually comes with practice but is there any other way to roughly get an idea of approximate i.e. thought process ?

The reason I ask is that there are lots of such problems requring the same approach...
FOR EXAMPLE, Please dont solve
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From a group of 3 boys and 3 girls, 4 children are to be randomly selected. What is the probability that equal numbers of boys and girls will be selected?

A. 1/ 10 B. 4/9 C 1/2 D 3/5 E 2/3
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Richard has 3 green, 2 red, and 3 blue balls in a bag.He randomly picks 5 from the bag without replacement.What is the probability that of the 5 drawn balls, Richard has picked 1red, 2 green, and 2 blue balls?
8/28 9/28 10/28 10/18 11/18
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Regards,
Ash

Stuart Kovinsky wrote:

Kaunteya wrote:If a jury of 12 people is to be selected randomly from a pool of 15 potential jurors, and the jury pool consists of 2/3 men and 1/3 women, what is the probability that the jury will comprise at least 2/3 men?
a. 24/91
b. 5/91
c. 2/3
d. 67/91
e. 84/91

Very similiar to the last problem that I posted but still haven't figured out how to solve these. If someone can please guide that I would really appreciate it. If you could solve this one, could you please solve the other one I posted titled: Combination and Permutation Manhattan. Thanks again guys.

Kaunteya

If 2/3 are men, we have 10 men and 5 women.

We want to know the probability that at LEAST 2/3 of the people actually selected will be men. In other words, that at least 8 out of the 12 jury members will be men.

There are three scenarios in which this could happen:

8 men and 4 women;

9 men and 3 women; and

10 men and 2 women.

Let's see how many different ways we can make each of these occur.

There are 10 men total, so there are 10C8 different groups of 8 men. There are 5 women, so there are 5C4 different groups of 4 women.

Therefore, scenario 1 has 10C8 * 5C4 = 10!/8!2! * 5!/4!1! = 45 * 5 = 225 possible juries.

For scenario 2, we have 10C9 * 5C3 = 10!/9!1! * 5!/3!2! = 10 * 10 = 100 possible juries.

For scenario 3, we have 10C10 * 5C2 = 10!/10!0! * 5!/2!3! = 1 * 10 = 10 possible juries.

[remember, 0!=1]

Now, since this is a probability question, we we want to use the probability formula.

Probability = #desired outcomes / total # of possible outcomes.

We've already calculated the # of desired outcomes: 225 + 100 + 10 = 335 juries with at least 8 men on them.

The total # of possible outcomes is the total # of possible juries, which is simply 15C12 = 15!/12!3! = 15*14*13/3*2*1 = 5*7*13 = lots, so let's reduce instead!

So:

335/5*7*13 = 67/7*13 = 67/91

choose (d).

Let's also look at this question from a strategic guessing point of view.

2/3 of the jury pool are men. Let's eliminate (c) 2/3, because that's way too easy.

Now we have a big split among the remaining choices. (a) and (b) are both very small (less than 1/3) and (d) and (e) are both big (more than 2/3). Since 2/3 of the jury pool are men, does it make any sense that there would be a small probability that 2/3 of the actual jury will be men too? Of course not, so (a) and (b) don't really make sense. So, if we're guessing, choose (d) or (e). (Further, we predict that the answer should be a bit more than 2/3 - (e) really seems too big, so (d) looks like the best guess.)

Never discount the power of strategic guessing, especially on really time consuming questions!

duongthang wrote:

sanjana wrote:

Kaunteya wrote:If a jury of 12 people is to be selected randomly from a pool of 15 potential jurors, and the jury pool consists of 2/3 men and 1/3 women, what is the probability that the jury will comprise at least 2/3 men?
a. 24/91
b. 5/91
c. 2/3
d. 67/91
e. 84/91

Very similiar to the last problem that I posted but still haven't figured out how to solve these. If someone can please guide that I would really
appreciate it. If you could solve this one, could you please solve the other one I posted titled: Combination and Permutation Manhattan. Thanks again guys.

Kaunteya

To this problem I came up with the below solution,

As found out by Stuart, there are 10 men and 5 Women from which the committee of 12 members needs to be picked.
Now the committee needs to be have atleast 2/3 Men,i.e atleast 8 men.
Thats P(M>=8) = 1-P(M<8)
=1-P(M=7)
=1-(10C7*5C5)15C12
=67/91,which is the correct Answer.

Stuart,

I have a question for you here..
For all problems that Involve Atleast or Atmost I almost always use the Rule P(A) = 1-P(A') and so far I have never got the wrong answer.
Is this the right approach or am I missing something?

Look forward to your valuable advice!

Thanks,
Sanjana

you are wrong. P(M>=8) is not equal to 1-P(M=7)

but P(M>=8) = 1- P(M=7)-P(M=6)-P(M=5)-P(M=4)-P(M=3)-P(M=2)-P(M=1)-P(M=0)

so the approach "one minus"here is not good