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#1 2006-10-15 16:15:17

eureka
Member
Registered: 2006-10-15
Posts: 1

Probability ... help urgently...

There are 5 men and 3 women in a room. Given probability for men with beard is 1/10.

a) find the probability that there are three men with beard and two with no beard.
(Do i need to count three men with beard X two men with not beard? or only count for three men with beard?)

b)if two people are selected randomly from the people in the room, what is the probabiliy that both of them are men with beard?

Any idea?

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#2 2006-10-15 21:48:57

mathsyperson
Moderator
Registered: 2005-06-22
Posts: 4,900

Re: Probability ... help urgently...

a)

If the first three men have beards and the last two don't, the probability is like this:
There is a 1/10 chance of having a beard, so the first 3 men each have a 1/10 chance of having a beard. Similarly, the last two men have a 9/10 chance of not having a beard. So the total probability is (1/10)³ * (9/10)² = 81/100000.

However, the beards can be on any combination of the men, rather than having to be on the first 3. There are 5!/(2!*3!) = 10 ways of arranging 3 beards on 5 men, so the above probability is multiplied by 10.

So your final answer is 81/100000 * 10 = 81/10000.

b)

This question has 2 parts. First you need to pick two men out of the group of people, then they both need to have beards.

Initially, there are 8 people in the group, 5 of which are men. So the probability of picking a man is 5/8. After you pick the first man, there are only 7 people left and 4 of those are men. So the probability of picking a 2nd man is 4/7. Multiplying these gives the probability of picking 2 men as 5/14.

In addition, both men need beards. The chance of a man having a beard is 1/10, so the chance of them both having beards is (1/10)² = 1/100. Combining this with the probability that you picked 2 men gives your final answer as 5/1400 = 1/280.


Why did the vector cross the road?
It wanted to be normal.

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#3 2006-10-18 22:04:19

Harish
Guest

Re: Probability ... help urgently...

Hi,
I do agree with the answer sent by mathsyperson.
But I think it can be done much simpler by using binomial distribution.
1)probability of three men with beards among five men is = 5C3*(1/10)^3*(9/10)^2 = 81/10000.
2) probablity of selecting two men with beards among eight people = probablity of selecting two men among eight people * probablity of selecting two men with beards among five men = (5C2/8C2)*[(2C2)*(1/10)^2*(9/10)^0]which turns out to be 1/280.

#4 2006-10-18 23:21:47

mathsyperson
Moderator
Registered: 2005-06-22
Posts: 4,900

Re: Probability ... help urgently...

That's the same method that I used. Mine just looks longer because I was padding it out with explanations and stuff.


Why did the vector cross the road?
It wanted to be normal.

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