Practice Probability for CAT

Probability questions make up an important part of the CAT Quantitative section. In this post you will see a number of practice problems modeled after actual CAT questions. Detailed solutions follow. Be careful though; some of these problems are very tricky!

CAT Probability practice

Probability Practice Problems

Problem 1

Two fair twelve-sided dice are thrown. What is the probability of getting a sum of 15?

  A. 1/16

  B. 5/72

  C. 1/12

  D. 7/24

Problem 2

In a certain precinct, 40% of the population identifies with the Red Party, 30% with the Blue Party, and the rest have no party affiliation. Of the Red Party, 50% are male; of the Blue Party, 60% are male; and of independents, 30% are male. What is the probability that a randomly selected female voter is neither Red Party nor Blue Party?

  A. 20/41

  B. 23/49

  C. 19/52

  D. 21/53

Problem 3

There are two bags of marbles on the table. The first bag has 6 blue and 3 yellow marbles, and the second bag has 5 blue and 3 yellow marbles. One bag is selected at random and two marbles chosen from the bag, one after the other without replacement. What is the probability that both marbles are yellow?

  A. 5/56

  B. 5/28

  C. 7/54

  D. 8/61

Problem 4

Two players take turns trying to eliminate the other in a game of chance. Player A has a 40% chance of eliminating Player B each turn. Player B has a 30% chance of eliminating Player A each turn. The game continues until one player is eliminated. If A goes first, what is the probability that A wins the game?

  A. 19/29

  B. 19/30

  C. 20/29

  D. 23/30

Problem 5

An ATM PIN number consists of 4 digits (0-9). What is the probability that a randomly assigned ATM PIN number has digits that add to 9?

  A. 103/10000

  B. 31/1000

  C. 11/500

  D. 21/1000

Problem 6

What is the probability that a 4 digit number that is a multiple of 11 will also be a multiple of 7?

  A. 116/819

  B. 1/7

  C. 117/818

  D. 13/90

Problem 7

In the card game of Poker, each player is dealt five cards from a standard pack of 52 cards. (In a standard deck, there are four different suits, and each suit contains 13 different denominations.) A three-of-a-kind is a Poker hand in which three cards have the same denomination while the other two are different denominations. What is the chance that you are dealt a three-of-a-kind on your first deal?

  A. 88/4165

  B. 97/6513

  C. 13/1035

  D. 15/3217

Problem 8

In how many ways can 5 distinct marbles be placed into 4 cups so that there is at least one marble in each cup?

  A. 220

  B. 240

  C. 360

  D. 480

Solutions

Are you ready to check your work? First of all, here’s the list of correct answers.

  1. B
  2. D
  3. A
  4. C
  5. C
  6. B
  7. A
  8. D

Solution to Problem 1

First, there are 122 = 144 possibilities for the outcome of rolling two 12-sided dice. Out of these, exactly 10 combinations result in a sum of 15:

3 + 12, 4 + 11, 5 + 10, 6 + 9, 7 + 8, 8 + 7, 9 + 6, 10 + 5, 11 + 4, and 12 + 3.

Thus the probability is 10/144 = 5/72.

Solution to Problem 2

Let D, R, and I be the events that a randomly selected voter is Red Party, Blue Party, and independent, respectively. Furthermore, let M and F be the events that the voter is male and female, respectively.

This problem asks for P(I | F), that is, the probability that a randomly selected voter is independent, given that the voter is female.

What we are given is: P(D) = 0.4, P(R) = 0.3, and P(I) = 1 – 0.4 – 0.3 = 0.3.

Moreover, we have the conditional probabilities, P(F | D) = 0.5, P(F | R) = 0.4, and P(F | I) = 0.7. (These are the complementary probabilities based on the given information about males.)

Now use Bayes’ Theorem.

BayesProblemSolution

Solution to Problem 3

First work out the probabilities for each bag separately.

Bag 1: B = 6 and Y = 3. The probability of getting a yellow first is 3/9. Then, there is one less yellow marble, so the chance of getting a second yellow would be 2/8. Multiply the probabilities to get 1/12.

Bag 2: B = 5 and Y = 3. The same method yields: 3/8 × 2/7 = 3/28.

Next, scale each probability by 1/2 to account for the fact that one of the two bags was chosen at random. Add the resulting fractions.

So, 1/24 + 3/56 = 15/168 = 5/56.

Solution to Problem 4

The key to solving this problem is to break it down into the rounds in which Player A might win. Either Player A eliminates the opponent (0.4 chance) or he/she does not (0.6 chance). In the event that Player A does not eliminate B in a given round, then we must also factor in the chance that Player B does not eliminate A on his/her turn (1 – 0.3 = 0.7 chance).

The following table summarizes the results in the first four rounds.

RoundProbability of Player A winning in this roundPartially simplified expression
10.40.4
2(0.6)(0.7)(0.4)(0.42)(0.4)
3(0.6)(0.7)(0.6)(0.7)(0.4)(0.42)2(0.4)
4(0.6)(0.7)(0.6)(0.7)(0.6)(0.7)(0.4)(0.42)3(0.4)

Notice that the probability of A winning in round n is exactly (0.4) × (0.42)n – 1. Add up all of these individual terms to find the probability of winning overall.

P(A wins) = Σn ≥ 1 (0.4) × (0.42)n – 1

The sum can be computed using the formula for a sum of a Geometric Series.

geom_series_example

Solution to Problem 5

First note that there are a total of 104 = 10000 possible PIN numbers. Assume that each combination is equally likely.

Now we just need to compute the number of 4-digit combinations that add to 9. This is an example of a “ball-and-bin” problem (see Permutations and Combinations for CAT for a quick review).

The number 9 is the sum of nine ones. Imagine distributing these nine ones into four positions, corresponding to the four digits in each ATM. For example, we could place two ones in the first position, none in the second, six in the third, and one in the last position:

1+1 | | 1+1+1+1+1+1 | 1

This corresponds to the PIN number 2061.

Thus, with the 9 ones being “balls” and the 4 digit positions being the “bins,” the total number of ways to do this is 12C3.

12 C 3

Therefore, the answer is 220/10000 = 11/500

Solution to Problem 6

Let’s make a careful count of the multiples. The first 4-digit number, 1000, is not a multiple of 11. However, because 1000 ÷ 11 = 90.9, we know that 11 × 91 = 1001 is the first multiple of 11 to consider.

Then since 9999 ÷ 11 = 909, we can list all 4-digit multiples of 11:

11 × 91, 11 × 92, 11 × 93, …, 11 × 908, 11 × 909

There are 909 – 91 + 1 = 819 of these.

Next, we must locate all multiples of 7 within this list. Because 11 and 7 are coprime, that is, they have no common factors, the trick is to look for multiples of 7 in the numbers 91, 92, 93, …, 908, 909.

Now 91 = 7 × 13 happens to be a multiple of 7. The next one is 98, and so on. Check 909 by dividing: 909 ÷ 7 = 129.86. Therefore, 7 × 129 = 903 is the final multiple of 7 in the list.

So that gives a total of 129 – 13 + 1 = 117 multiples of 7 within the list of multiples of 11.

Thus, the fraction of multiples of 11 that are also multiples of 7 is equal to: 117/819 = 1/7.

Solution to Problem 7

This is a popular kind of problem which requires a careful step-by-step counting procedure.

First of all, the total number of five-card hands is: 52C5. This number is rather large:

52 C 5

Next, we must count the hands that qualify as three-of-a-kind.

  1. Choose one of the 13 denominations to be the repeated one: 13 choices.
  2. Of that denomination, pick 3 of the 4 cards to be in your hand: 4C3 = 4 combinations.
  3. Of the remaining 52 – 4 = 48 cards, choose two to fill out the hand: 48C2 = (48 · 47)/(2 · 1) = 1128. However, we have actually over-counted in this step! From 1128, take away the choices that represent having the same denominations in those final two cards. Remember, there are now 12 denominations to work with, so that’s 12 × 4C2 = 12 × 6 = 72. So that leaves 1128 – 72 = 1056.

Therefore, upon multiplying the number of choices at each step above, we obtain: 13 × 4 × 1056 = 54,912.

Finally, let’s put it all together. Divide the number of desirable hands by the total hands and reduce the fraction.

reduce_large_fraction

Solution to Problem 8

Ok, so this one isn’t really a probability problem. But it does display the kinds of subtle counting arguments that you often must perform when computing probabilities.

It helps to think about the situation before attempting to set up any formulas. The key is to first place 4 of the marbles into the 4 cups. There are 5C4 = 5C1 = 5 ways to pick these 4 marbles. Then there are 4! = 24 arrangements into 4 cups.

So far, we have 5 × 24 = 120 possibilities.

Finally, the fifth marble has to go into one of the 4 cups. There are clearly 4 choices for this step.

That gives 120 × 4 = 480 possibilities.

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