Probability question

[ritika_j]

Q Barbara has 8 shirts and 9 pants. How many clothing combinations does barbara have, if she doesnt wear 2 specific shirts with 3 specific pants?

I did it this way 8 shirts -2 shirts=6
9 pants -3 pants=6

so number of comibinations=6*6=36
but the answer is 66
9*8=72-3*2=66

please explain the difference btw two methods... so that it becomes clear.

[pendyal]

Originally Posted by ritika_j

Q Barbara has 8 shirts and 9 pants. How many clothing combinations does barbara have, if she doesnt wear 2 specific shirts with 3 specific pants?

I did it this way 8 shirts -2 shirts=6
9 pants -3 pants=6

so number of comibinations=6*6=36
but the answer is 66
9*8=72-3*2=66

please explain the difference btw two methods... so that it becomes clear.

u fogot to take into conideration the two cases where barbara wears one of the 2 shirts and one of the remaining 6 pants;and where she wears one of the 3 pants and one of the remianing 6 shirts.

so to the figure u got(36) add 2*6 and 3*6 which will give u 66.

bye..

[ritika_j]

Q The probability of pulling a black ball out of a glass jar is 1/X. The probability of pulling a black ball out of a glass jar and breaking the jar is 1/Y. What is the probability of breaking the jar?


plz solve this question...

[billybolimeeaaaw]

Originally Posted by ritika_j

In how many ways can the letter of the word "daughter" be arranged such that no vowels are together?

answer this question with explanation

consonants:d,g,h,t,r
vowels:a,u,e

_ _ _ _ _ _ _ _
1 2 3 4 5 6 7 8

no. of ways in which 5 consonants can be arranged: 5!=120
no. of ways in which 3 vowels can b arranged in the 6 spaces between them= 6P3=120
so answer=120*120=14400

elegant eh???

[supervish]

Originally Posted by ritika_j

Q The probability of pulling a black ball out of a glass jar is 1/X. The probability of pulling a black ball out of a glass jar and breaking the jar is 1/Y. What is the probability of breaking the jar?

I dunno if its really as simple as it seems, but I do feel like saying DUH !

event : pulling a black ball out = b

event : breakin the jar = j

p( ball pull AND jar phoda ) = p(b)*p(j) = 1/Y

to P(j) = (1/Y) / (1/X) = X/Y.

Aur kuch, or did i hagofy ?

[pendyal]

me thinks vish is rite but me also not too sure.
cud some1 provide the answer with absolut authority.

bye..

[ritika_j]

vish ur answer is right ... that is the official anwer.

[capricious]

cud some1 provide the answer with absolut authority.

P(breaking jar and taking out ball)=P(breaking jar) x P(taking out ball)
1/y=p*1/x
=> p=x/y

There is no ambiguity left.

[kapil.khurana]

No need to do C(64,2) because even in numerator, we are selecting first square followed by second square.

[sumit82]

Can someone help me with this, I can;t get where exactly I am going wrong.....


Q. Kurt, a painter, has 9 jars of paint 4 of which are yellow, 2 are red and the remaining jars are brown. Kurt will combine 3 jars of paint into a new container to make a new color which he will name according to the following conditions:

Brun Y if the paint contains 2 jars of brown paint and no yellow
Brun X if the paint contains 3 jars of brown paint
Jaune X if the paint contains at least 2 jars of yellow
Jaune Y if the paint contains exactly 1 jar of yellow

What is the probability that the new color will be jaune?

A) 5/42
B) 37/42
C) 1/21
D) 4/9
E)5/9

I did.........

4/9*5/8*4/7 + 4/9*3/8*5/7 + 4/9*3/8*2/7
(Jaune Y) + (Jaune X) + (Jaune X)

= 41/126

??????