mirhaque wrote:

These are two probability questions from GMAT club probability course. Can anybody explain these with concepts for a beginner? Thanks.

1. If the coin is tossed 5 times, what is the probability that at least 3 out of 5 times it will show heads? (Binominal Distribution)

2. There are 2 green, 3 red, and 2 blue balls in a box. 4 are drawn at random without replacement. What is the probability that of the 4 drawn balls two are red, 1 is green, and 1 is blue? (Hypergeometric Distribution)

1. If you toss a coin 5 times, then you could get head 0, 1, 2, 3, 4, 5 times. The probabilities of these events are:

0 times: each time you get NOT head, (suppose that head shows with prob. 0<p<1), so the probability of this event = (1-p)*(1-p)*(1-p)*(1-p)*(1-p). This is so because outcomes of tossing are independent and each outcome has probability of 1-p.

1 times: you have 5 options - the coin showed head in the first tossing, int the second, int the third, ... For each option, you compute the probability again using that independence rule - p*(1-p)^4. 5 times this leads to an answer 5*p*(1-p)^4.

..... etc.

2. This problem resembles that with cards in a deck. There were 52 cards (7 balls in this case), 13 hearts and 39 non-hearts (in this case - 3 red, 2 green, and 2 blue from total 7 balls). Then you start taking without replacement: you should just remeber the formula. Or, if you know that formulas are hard to remember for you, you can use the following method:

P(r = 2, g = 1, b = 1) = P(first is green)*P(r = 2, g = 1, b = 1| first is green) + P(first is red)*P(r = 2, g = 1, b = 1| first is red) + P(first is blue)*P(r = 2, g = 1, b = 1| first is blue).

P(first is red) = |red|/|total|, similarly for green and blue.

P(r = 2, g = 1, b = 1| first is green) = P(r = 2, g = 1, b = 1| first is green, second is blue)*P(second is blue| first is green) + P(r = 2, g = 1, b = 1| first is green, second is red)*P(second is red| first is green).

and so on. So, you clearly should remeber formulas to easily(and, it's important, quickly) calculate it all...