Сегодня мы поговорим задачах, которые достаточно часто встречаются в математической части теста GMAT. Это задачи по поиску вероятностей.  Эффективное использование формул для вероятности поможет Вам сэкономить время при решении задач и, таким образом, улучшит общий результат.

We are talking about probability problems today. They can take less time if you know some formulae.
The probability that an event A occurs, denoted by P(A) , is a number between 0 and 1, inclusive.
The most useful properties for probability problems you can find here:

1.    P(A&B) = P(A)*P(B) or, the probability of  A happening and B happening is the same as the probability of A happening times the probability of B happening.
2.    P(A or B) = P(A)+P(B) or, the probability of only A happening or only B happening is the same as the probability of A happening plus the probability of B happening.
3.    P(A) =1 – P(not A) or, the probability of an event occurring is always the same as one minus the probability of the event not occurring .

1.    Two dices are tossed. What is the probability that we’ll get 5 on both?
(A) 1/36
(B) 1/6
(C) 1/3
(D) 35/36
(E) 1/18

Probability of getting 5 on the first dice is 1/6 (1 side suits us of 6 possible).
Probability of getting 5 on the second dice is the same: 1/6.
We need both events to occur, so P = 1/6 * 1/6 = 1/36. Answer A.

2.     If x is to be chosen at random from the set {1, 2, 3, 4} and y is to be chosen at random from the set {5, 6, 7}, what is the probability that xy will be even?
(A)1/6
(B)1/3
(C)1/2
(D)2/3
(E)5/6
1) P(x will be even) = 2/4 = ½=P(x will be odd), P(y will be even)= 1/3, P(y will be odd) =2/3.
Therefore, P(xy will be even) = P(x –even, y-odd)+P(x- odd, y-even) +P(x,y –even) =1/2 * 2/3 + ½ * 1/3 + ½ * 1/3 = 2/3. See, to count P(x –even, y-odd), we multiple the probability for  x to be even AND for y to be odd. And as for total probability, we add, because the situation when x –even, y-odd  OR x- odd, y-even OR x,y –even suit us.
2) this problem can be also solved using subtraction rule - the probability of an event occurring is always the same as one minus the probability of the event not occurring. The situation that does not sit us is when both x and y are odd. We can find its probability: P(x,y- odd)  =1/2* 2/3 = 1/3. The, the probability of our event is: P(xy  - even)= 1- P(x,y – odd)= 1 -1/3 = 2/3. As you see, we’ve got the same answer.
Therefore, D  is the best answer.

3.    The hold of a fishing boat contains only cod, haddock, and halibut. If a fish is selected at random from the hold, what is the probability that it will be a halibut or a haddock?
(1) There are twice as many halibut as cod in the hold, and twice as many haddock as halibut.
(2) Cod account for 1/7 of the fish by number in the hold.

(A) Statement (1) alone is sufficient, but statement (2) alone is not sufficient.
(B) Statement (2) alone is sufficient, but statement (1) alone is not sufficient.
(C) BOTH statements TOGETHER are sufficient, but NEITHER statement ALONE is sufficient.
(D) Each statement ALONE is sufficient.
(E) Statements (1) and (2) TOGETHER are NOT sufficient.

(1) alone is sufficient because it means that if there is one cod, there are two halibut and four haddock, so halibut and haddock account for 6/7 of the fish in the hold, giving a probability of 6/7 of selecting one of these fish. (2) alone is sufficient, because the probability of selecting a cod is the inverse of the probability of selecting one of the other two fish, or 6/7. Thus, the answer is D.

Resume. In this lesson we talked about problems, in which we have to calculate the probability of the event. Such problems appear quite often on the test. Using of simple rules gives you right solution and opportunity to save some time on your GMAT.

Material prepared by Ksenia Zueva,
GMAT consultant at MBA Strategy