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Probability basics

If i have to chose two specific numbers from a set of 5 numbers:
Eg: 2 and 3 from set {1,2,3,4,5}

I understand the basic counting principle gives 2/5. (2 out of 5)
Using Combinations Will it be 2c2/5c2? or 2c2/5p2?

by shashwats

Wed Jun 12, 2013 7:01 am
Forum: Problem Solving
Topic: Probability basics
Replies: 2
Views: 1435

But does it not say "At least one of the colour and model" which means even RED and Large case should be included?

by shashwats

Sun Jun 09, 2013 7:44 am
Forum: Problem Solving
Topic: Some toys include large, middle, and small model with r
Replies: 3
Views: 2033

What about ?
0.5% of x + 0.8% of y= 0.75% of (x+y)
solving we get y= 5x.
hence y proportion is 5/6

by shashwats

Mon May 13, 2013 11:46 pm
Forum: Problem Solving
Topic: Problem Solving
Replies: 6
Views: 1575

Problem Solving

Q:Each of the products produced yesterday was checked by x or y. 0.5% of the products checked by x are defective and 0.8% of the products checked by y are defective. If the total defective rate of all the products checked by x and y is 0.75%, what fraction of the products was checked by y?

by shashwats

Mon May 13, 2013 11:12 pm
Forum: Problem Solving
Topic: Problem Solving
Replies: 6
Views: 1575

Probability

If there are 10 balls, 3 red, 5 blue and 2 green. I pick two without replacement, what is the probability that both are red?

I find two ways of doing this, clearly one is wrong!

1) 3/10 * 2/9

2) (3C1 *2C1)/10C2

Can someone please help me understand this!

by shashwats

Thu May 02, 2013 12:13 pm
Forum: Data Sufficiency
Topic: Probability
Replies: 3
Views: 1229

Thanks anju@gurome.
I had taken 21C3 instead of 20C3, hence the error!

by shashwats

Mon Apr 29, 2013 6:31 am
Forum: Problem Solving
Topic: Probability
Replies: 9
Views: 3285

Averages

A certain mixture of nuts consists of 4 parts almonds to 1 part walnuts, by weight. By removing some of the almonds, the remained nuts consist 75 percent of almonds, by weight. What percent of the almonds were removed?
(A) 10%
(B) 12.5%
(C) 20%
(D) 25%
(E) 30%

by shashwats

Mon Apr 29, 2013 4:55 am
Forum: Problem Solving
Topic: Averages
Replies: 4
Views: 1728

Probability

If 5 numbers are to be selected from integers from 1 to 20, inclusive, without replacement, what is the probability that both 5 and 15 will be selected?
(A) 1/57
(B) 1/19
(C) 3/19
(D) 4/57
(E) 2/19

What I thought of was:
(2C2 * 19C3)/ 21C5

by shashwats

Mon Apr 29, 2013 4:43 am
Forum: Problem Solving
Topic: Probability
Replies: 9
Views: 3285