A has to pick a ball from a bag with 3 red & 3 blue balls.

Question

A has to pick a ball from a bag with 3 red & 3 blue balls. B has to pick a ball from another bag with 3 red, 3 blue & 3 green balls. C has to pick a ball from another bag with 3 red, 3 blue, 3 yellow & 3 green balls. 1. What is the probability that A or B get a red ball, & not C? 2. What is the probability that A and B get a red ball, & not C? PS: This is just a random question. :idea:

My initial thought process was : All 3 are independent events. So, p1 = P(A or B) x P(!C) = ( P(A) + P(B) - P(A)P(B) ) x P(!C) = (3/6 + 3/9 - 3/6 * 3/9) * 9/12 p2 = P(A and B) x P(!C) = ( P(A)P(B) ) x P(!C) = (3/6 * 3/9) * 9/12 Is my approach correct?

mbafall2011 · Answer

Seems right. I would go about it the same way.

InSearch · Answer

My take:
We are dealing with probabilities, not sets. P(X or Y) = P (X) + P(Y).

P(A gets a red ball) = 1/2. 
P(B gets a red ball) = 1/3.
P(C does not get a red ball) = 3/4

A1: (1/2+1/3)*3/4 = 5/8
A2: 1/2 * 1/3 * 3/4 = 1/8.

mbafall2011 · Answer

Sorry my bad! i didnt see the -P(AandB) in the original post. This makes sense. Does any one have the correct answer?

idiot · Answer

i believe the generic formula is : P(A OR B) = P(A) + P(B) - P(A/B) :idea:

where P(A/B) is the probability of occurrence of A given B has occurred ,i.e, the probability of occurrence of A & B. (or use P(B/A) instead) Now, For independent events, P(A/B) = P(A)*P(B) (as in this example) For dependent events, P(A/B) != P(A)*P(B) (e.g: drawing cards without replacement from a deck) For mutually exclusive events P(A/B) = 0 (as A & B cant occur together) lets take a simple example: I roll a fair dice. What's the probability that i get a prime or an odd number? Solution: (i) counting method p = favorable outcomes{1,2,3,5} / total outcomes = 4 / 6 . (ii) individual probabilities method: p(prime){2,3,5} = 3 / 6 p(odd){1,3,5} = 3 / 6 p(prime and odd){3,5} = 2 / 6 so, p(prime or odd) = 3/6 + 3/6 - 2/6 = 4 / 6 . My understanding is that set theory & probability theory are, in a way, analogous. Any inputs? Anybody?

Bunuel · Answer

Yes, your approach is correct.

1. What is the probability that A or B get a red ball, & not C?

NOTE:  Probability of A OR B.
If Events A and B are independent, the probability that either Event A or Event B occurs is:
 P(A or B) = P(A) + P(B) - P(A and B)

When we say "A or B occurs" we include three possibilities:
A occurs and B does not occur;
B occurs and A does not occur;
Both A and B occur.

P(AorB&notC)=P(AorB)*P(notC)=(P(A) + P(B) - P(A and B))*P(notC)= 
 =(\frac{3}{6}+\frac{3}{9}-\frac{3}{6}*\frac{3}{9})*\frac{9}{12}=\frac{1}{2}

2. What is the probability that A and B get a red ball, & not C?

NOTE:  Probability of A and B.
When two events are independent, the probability of both occurring is the product of the probabilities of the individual events:  P(A and B) = P(A)*P(B)

P(A&B&notC)=P(A)*P(B)*P(notC)=\frac{3}{6}*\frac{3}{9}*\frac{9}{12}=\frac{1}{8}

pappueshwar · Answer

HI bunuel,

please let me know as to how you arrived at 9/12 for p(not C)

Bunuel · Answer

Sure.

Given: C has to pick a ball from a bag with 3 red, 3 blue, 3 yellow & 3 green balls. What is the probability that C will not get a red ball?

Since there are total of 3+3+3+3+=12 balls out of which 9 balls are not red then the probability of picking the ball other than red is 9/12.

Hope it's clear.

sfeng · Answer

Hi Bunuel,

I did it the same way as InSearch - not clear to me why it's wrong.

Why is the answer for 1 not 5/8?
ie. for your calculation of P(A or B), why do you subtract the P(A and B)? 
The question says "from another bag", which to me indicates that A and B are mutually exclusive.
Also, how did you get 3/6 for P(A and B) in question 1? (And why/how is it something different in question 2?)

Thank you!

rrsnathan · Answer

We are dealing with probabilities,

P(A gets a red ball) = 1/2.
P(B gets a red ball) = 1/3.
P(C does not get a red ball) = 3/4

A1: (1/2+1/3)*3/4 = 5/8
A2: 1/2 * 1/3 * 3/4 = 1/8.

Is this approach is wrong????
Please help me

Asifpirlo · Answer

.................................
your procedure is applicable for only one bag.
if there are different bags then there must be different events or independent events, and for every independent event we have to count p(A n B) = p(A) . p(B)
so you have to minus p(A n B).
From your calculation , (1/2 + 1/3 - 1/2 * 1/3) * 3/4 = 1/2 (just include one minus, because of independent events)

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A has to pick a ball from a bag with 3 red & 3 blue balls.

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