**Probability questions in the UMAT: you’re probably reading this sentence right now…**

*Your UMAT questions explained by the team at MedEntry as part of our new section UMAT test tactics and preparation.*

We know what probability means, but what is its formal definition? Let’s use our everyday logic to define it. If there is no chance that an event will occur, then its probability of occurring should be 0. On the other extreme, if an event is certain to occur, then its probability of occurring should be 100%, or 1. Hence, we can deduce that probability will be a number between 0 and 1, inclusive. But what kind of number? Suppose your favourite actor has a 1 in 3 chance of winning the Oscar for best actor. This can be measured by forming the fraction 1/3. Hence, a probability is a fraction where the top is the **number of ways an event can** occur and the bottom is the **total number of possible events**:

P= (Number of ways an event can occur)/(Number of total possible events)

**Example UMAT style question:: **

**Flipping a coin**

What’s the probability of getting heads when flipping a coin?

There is only one way to get heads in a coin toss. Hence, the top of the probability fraction is 1. There are two possible results: heads or tails. Forming the probability fraction gives 1/2.

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** Example UMAT style question:** Tossing a die

What’s the probability of getting a 3 when tossing a die?

A die (a cube) has six faces, numbered 1 through 6. There is only one way to get a 3. Hence, the top of the fraction is 1. There are 6 possible results: 1, 2, 3, 4, 5, and 6. Forming the probability fraction gives 1/6.

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** Example UMAT style question**: Drawing a card

**from a deck**

What’s the probability of getting a king when drawing a card from a deck of cards?

A deck of cards has four kings, so mere are 4 ways to get a king. Hence, the top of the fraction is 4. There are 52 total cards in a deck. Forming the probability fraction gives 4/52, which reduces to 1/13. Hence, there is 1 chance in 13 of getting a king.

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** Example UMAT style question**: Drawing marbles from a bowl

What’s the probability of drawing a blue marble from a bowl containing 4 red marbles, 5 blue marbles, and 5 green marbles?

There are five ways of drawing a blue marble. Hence, the top of the fraction is 5. There are 14 (= 4 + 5 + 5) possible results. Forming the probability fraction gives 5/14.

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** Example UMAT style question:**: Drawing marbles from a bowl (second drawing)

What’s the probability of drawing a red marble from the same bowl, given that the first marble drawn was blue and was not placed back in the bowl?

There are four ways of drawing a red marble. Hence, the top of the fraction is 4. Since the blue marble from the first drawing was not replaced, mere are only 4 blue marbles remaining. Hence, there are 13 (=4 + 4 + 5) possible results. Forming the probability fraction gives 4/13.

**Consecutive Probabilities**

What’s the probability of getting heads twice in a row when flipping a coin twice? Previously we calculated the probability for the first flip to be 1/2. Since the second flip is not affected by the first (these are called mutually exclusive events), its probability is also 1/2. Forming the product yields the probability of two heads in a row: (1/2) * (1/2) = 1/4

What’s the probability of drawing a blue marble and then a red marble from a bowl containing 4 red marbles, 5 blue marbles, and 5 green marbles? (Assume that the marbles are not replaced after being selected.) As calculated before, there is a 5/14 likelihood of selecting a blue marble first and a 4/13 likelihood of selecting a red marble second. Forming the product yields the probability of a blue marble immediately followed by a red marble: (5/14) * (4/13) = 20/182 = 10/91

These two examples can be generalized into the following rule for calculating consecutive probabilities:

**To calculate consecutive probabilities, multiply the individual probabilities.**

- This rule applies to two, three, or any number of consecutive probabilities.

**Either-Or Probabilities**

What’s the probability of getting either heads or tails when flipping a coin once? Since the only possible outcomes are heads or tails, we expect the probability to 100%, or 1: 1/2 + 1/2 = 1. Note that the events heads and tails are mutually exclusive. That is, if heads occurs, then tails cannot (and vice versa).

What’s the probability of drawing a red marble or a green marble from a bowl containing 4 red marbles, 5 blue marbles, and 5 green marbles? There are 4 red marbles out of 14 total marbles. So the probability of selecting a red marble is 4/14 = 2/7. Similarly, the probability of selecting a green marble is 5/14. So the probability of selecting a red or green marble is (2/7) + (5/14) = 9/14. Note again that the events are mutually exclusive. For instance, if a red marble is selected, then neither a blue marble nor a green marble is selected.

These two examples can be generalized into the following rule for calculating either-or probabilities:

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**To calculate either-or probabilities, add the individual probabilities (only if the events are mutually exclusive).**

The probabilities in the two immediately preceding examples can be calculated more naturally by adding up the events that occur and then dividing by the total number of possible events. For the coin example, we get 2 events (heads or tails) divided by the total number of possible events, 2 (heads and tails): 2/2=1. For the marble example, we get 9 (= 4 + 5) ways the event can occur divided by 14 (= 4 + 5 + 5) possible events: 9/14.

If it’s more natural to calculate the either-or probabilities above by adding up the events that occur and then dividing by the total number of possible events, why did we introduce a second way of calculating the probabilities? Because in some cases, you may have to add the individual probabilities. For example, you may be given the individual probabilities of two mutually exclusive events and be asked for the probability that either could occur. You now know to simply add their individual probabilities.

For a more extensive data-base of UMAT questions and more detailed UMAT advice see MedEntry UMAT preparation.