GMAT

The types of events that we have discussed so far are all independent events. By independent we mean that the first event does not affect the probability of the second event. Coin tosses are independent. They cannot affect each other's probabilities; the probability of each toss is independent of a previous toss and will always be 1/2. Separate drawings from a deck of cards are independent events if you put the cards back. An example of a dependent event, one in which the probability of the second event is affected by the first, is drawing a card from a deck but not returning it. By not returning the card, you've decreased the number of cards in the deck by 1, and you've decreased the number of whatever kind of card you drew. If you draw an ace of spades, there are 1 fewer aces and 1 fewer spades. This affects our simple probability: (number of favorable outcomes)/ (total number of outcomes. This type of probability is formulated as follows:

Example 6

If someone draws a card at random from a deck and then, without replacing the first card, draws a second card, what is the probability that both cards will be aces?

Solution
Event A is that the first card is an ace. Since 4 of the 52 cards are aces, P(A) = 4/52 = 1/13. Given that the first card is an ace, what is the probability that the second card will be an ace as well? Of the 51 remaining cards, 3 are aces. Therefore, P(B|A) = 3/51 = 1/17, and the probability of A and B is 1/13 × 1/17 = 1/221. The same reasoning is applied to marbles in a jar.

Example 7

If there are 30 red and blue marbles in a jar, and the ratio of red to blue marbles is 2:3, what is the probability that, drawing twice, you will select two red marbles if you return the marbles after each draw?

Solution
First, let's determine the number of red and blue marbles respectively. The ratio 2:3 tells us that the total of 30 marbles must be broken into 5 groups of 6 marbles, each with 2 groups of red marbles and 3 groups of blue marbles. Setting up the equation 2x + 3x = 5x =30 employs the same reasoning. Solving, we find that there are 12 red marbles and 18 blue marbles. We are asked to draw twice and return the marble after each draw. Therefore, the first draw does not affect the probability of the second draw. We return the marble after the draw, and therefore, we return the situation to the initial conditions before the second draw. Nothing is altered in between draws, and therefore, the events are independent.

Now let's examine the probabilities. Drawing a red marble would be 12/30 = 2/5. The same is true for the second draw. Since we want two red marbles in a row, the question is really saying that we want a red marble on the first draw and a red marble on the second draw. The "and" means we should expect a lower probability than 2/5. Understanding that the "and" is implicit can help you eliminate choices d and e which are both too big. Therefore, our total probability is P(A and B) = P(A) ×. P(B) = 2/5 × 2/5 = 4/25.

Now consider the same question with the condition that you do not return the marbles after each draw. The probability of drawing a red marble on the first draw remains the same, 12/30 = 2/5. The second draw, however, is different. The initial conditions have been altered by the first draw. We now have only 29 marbles in the jar and only 11 red. Don't panic! We simply use those numbers to figure our new probability of drawing a red marble the second time, 11/29. The events are dependent and the total probability is P(A and B) = P(A) ×. P(B) = 2/5 × 11/29 = 132/870 = 22/145.

If you return every marble you select, the probability of drawing another marble is unaffected; the events are independent. If you do not return the marbles, the number of marbles is affected and therefore dependent.

The first jar contains 4 blue and 5 red marbles; the second basket contains 3 blue and 4 red marbles. One marble is randomly extracted from the first basket and put into the second. After that, a marble is extracted from the second basket. What is the probability that this marble is blue?

1/3
15/36
31/72
4/9
11/18

3. C

Among all possible scenarios there are two that suit us:

1. A blue marble is put into the second basket and then a blue marble is extracted from the second basket;

2. A red marble is put into the second basket and then a blue marble is extracted from the second basket.

The probability of the first scenario = probability that a blue marble is taken from the first basket × probability that a blue marble is then extracted from the second basket = 4/9 × 1/2 = 4/18.

The probability of the second scenario = probability that a red marble is taken from the first basket × probability that a blue marble is then extracted from the second basket = 5/9 × 3/8 = 15/72.

The probability of EITHER first OR second scenario = 4/18 + 15/72 = 31/72.

The important thing is to understand that the probability of drawing a blue marble from the second basket depends on what marble was put there. For example, in the first scenario the probability of drawing a blue marble from the second basket is 1/2 because after one blue marble was added to the second basket, the basket contains equal number of marbles of each color. This is an example of a question the uses an implicit OR statment for the probability.

2. After each throw of a die, the face that shows is marked with red color. What is the probability that after 6 throws all faces of the die will be marked red?

2. B

We have to find the probability that each of the die's faces will appear once in the 6 throws. There are no limitations on the result of the first throw. The second throw has to produce a result which is different from the result of the first throw. The probability of this is 5/6. The third throw has to produce a result which is different from the results of the first two throws. The probability of this is 4/6. Ultimately, we have to calculate 5/6 × 4/6 × 3/6 × 2/6 × 1/6 which is.