## Probability of Failure

Let’s revisit the original example:

*You have a single six-sided die. If you roll it once, what is the probability you come up with a 5?*

The probability is 1/6.

Now let’s bring some money into this. You’re in Vegas, and you’re going to win $500 if the die lands on 2. You have a 1/6 or 1 in 6 chance of success.

But what is your chance of failure? Since 5 out of the 6 outcomes would cause failure, you have a 5/6 probability of failure.

Now a different scenario. What if someone were to offer you $500 if, when you roll one die, you rolled **at least a 2? **Think about winning outcomes. If you roll a 2, 3, 4, 5 or 6, you would win. What is the probability of that happening? There are 5 ways to win out of 6, so the **probability of success** would be 5/6.

Here, solving for “at least” or “or” was fairly simple.

probability of rolling at least a 2

= (probability of rolling a 2)

+ (probability of rolling a 3)

+ (probability of rolling a 4)

+ (probability of rolling a 5)

+ (probability of rolling a 6)

= 1/6 + 1/6 + 1/6 + 1/6 + 1/6 = 5/6

By using the phrase “at least,” the problem created more ways to succeed than to fail. Notice that finding this probability required more calculations.

Instead of trying to find the probabilities of the ways to win, you can find the probability of failure and subtract it from 1. For rolling at least a 2, the only way NOT to win is to roll a 1. The probability of rolling a 1 is 1/6. If the probability of failure is 1/6, the probability of success must be 5/6.

As a formula:

P(success) + P(failure) = 1 ** ** P(success) = 1 – P(failure)

or

## Example

A bag contains 10 red marbles and 6 black marbles. If 3 marbles are pulled from the bag without replacement, what is the probability there will be at least 1 red marble?

(A) 1/28

(B) 5/168

(C) 3/8

(D) 27/28

(E) 167/168

### Solution

Finding the probability of success would be time consuming. For pulling just 1 red marble, it could be the first, second or third marble pulled: **R B B** or **B R B** or **B B R**

Pulling 2 red marbles also fulfills the “at least 1 red marble”: **R R B** or **R B R** or **B R R**

As does pulling 3 red marbles: **R R R**

So there are 7 possible desired outcomes. But calculating the total possible outcomes takes much more calculation.

Instead, calculate the probability of failure. The only failure is to draw no red marbles which means drawing all black marbles.

**B B B**

Probability of drawing a black marble on the first draw:

6/16 = 3/8

Probability of drawing a black marble on the second draw:

5/15 = 1/3

Probability of drawing a black marble on the third draw:

4/14 = 2/7

This is dependent probability, so multiply the probabilities.

\dfrac{3}{8} × \dfrac{1}{3} × \dfrac{2}{7} = \dfrac{1}{28}\\[2ex]Since the probability of pulling only black marbles is 1/28, the probability of drawing at least 1 red marble is 1 – 1/28 = 27/28.

Make sure you do this last step. A common trick on the SAT is to have you to use the result from one part of a question to get the real answer. Typically both results will be available as answer choices.

The correct answer is choice (D).

## Example

Two six-sided dice are rolled together. What is the probability that:

a) the sum of the dice will be at least 5?

b) the sum of the dice will not be 6?

### Solution

**a) **Notice the “at least” and think about what it means for this problem. Counting all the possible pairs of dice whose sum would be 5 or greater would take some time. Instead, figure out the probability of failure, which is a sum less than 5. So the sum can be 2, 3 or 4. It is easy to make a list of these outcomes.

(1,1) (1, 2) (1, 3) (2, 1) (2, 2)

(3, 1)

So there are 6 ways to come up with 4 or less. Each dice pair has a probability of 1/36, so together the probability of failure is 6/36 = 1/6.

The probability of success, rolling a sum that is at least 5, is

1 – 1/6 = 5/6.

**b) **Notice the “will not be.” Counting all the pairs of dice that don’t sum to 6 would be a long process. Instead, figure out the probability of failure, which is a sum equal to 6. It is easy to make a list of these outcomes.

(1, 5) (2, 4) (3, 3)

(5, 1) (4, 2)

So there are 5 ways to sum to 6. Each dice pair has a probability of 1/36, so together the probability of failure is 5/36.

The probability of success, rolling a sum that does not sum to 6, is

1 – 5/36 = 31/36.

## Example

A bag contains

rred marbles,

5r+ 4 white marbles and 3r+ 2 blue marbles. If there are only red, white and blue marbles in the bag, what is the probability of pulling a red or white marble?

### Solution

Don’t let the variables scare you. Use the same process, and remember to factor.

The total number of marbles in the bag is *r* + (5*r* + 4) + (3*r* + 2) = 9*r* + 6.

*Method 1** Working backwards*

The probability of pulling a red or white marble = 1 − probability of drawing a blue marble.

probability = 1 – \dfrac{blue \,marbles}{total \,marbles}

= 1 \,-\, \dfrac{3\textit{r} + 2}{9\textit{r} + 6} = 1 \,-\, \dfrac{3\textit{r} + 2}{3(3\textit{r} + 2)}

= 1 \,-\, \dfrac{1}{3} =\dfrac{2}{3}

*Method 2 **Straightforward*

probability = \dfrac{red + white \,marbles}{total \,marbles}

= \dfrac{\textit{r} + 5\textit{r} + 4}{9\textit{r} + 6} = \dfrac{6\textit{r} + 4}{9\textit{r} + 6}

= \dfrac{2(3\textit{r} + 2)}{3(3\textit{r} + 2)} = \dfrac{2}{3}

## Odds *(advanced topic)*

Finding odds has a slightly different meaning than finding probability. ** Odds** compare the number of desired to the number of undesired outcomes.

Though the SAT will probably not ask about odds, seeing how odds and probability are different can help you avoid making mistakes.

Compare the denominators.

**probability **= \dfrac{desired \,outcomes}{all \,possible \,outcomes}

**odds** = \dfrac{desired \,outcomes}{undesired \,outcomes}

= \dfrac{desired \,outcomes}{all \,outcomes \,-\, desired \,outcomes}

Compare the values of probabilities and odds in these examples.

## Example

A fair six-sided die is rolled once.

a) What is the probability of rolling a 5?

b) What are the odds of rolling a 5?

### Solution

**a)**

**probability** = \dfrac{1 \,desired \,outcome}{6 \,possible \,outcomes}

= \dfrac{1}{6}

**b)**

**odds** = \dfrac{1\, desired\, outcome}{6\, possible \,outcomes\,-\, 1 \,outcome}

= \dfrac{1}{5}

## Example

A deck of 52 cards contains cards numbered 1 through 13 in each of 4 different colors: green, yellow, blue and red.

a) What is the probability of drawing a card numbered ten?

b) What are the odds of drawing a card numbered ten?

### Solution

**a)**

**probability** = \dfrac{4 \,tens}{52 \,cards} = \dfrac{1}{13}

So the probability of drawing a ten is 1/13, or a 1 in 13 chance.

**b)**

**odds** = \dfrac{4 \,tens}{52 \,cards \,-\, 4\, tens} = \dfrac{4}{48} = \dfrac{1}{12}

The odds are 1 to 12 *against* drawing a ten.

## Example

A deck of 52 cards contains cards numbered 1 through 13 in each of 4 different colors: green, yellow, blue and red.

a) What is the probability of

notdrawing a blue card?b) What are the odds of

notdrawing a blue card?

### Solution

**a)**

The probability of not drawing a blue card is 1 – (the probability of drawing a blue card).

1 – (probability of drawing a blue card)

= 1 – \dfrac{13 \,blue \,cards}{52\, cards} = \dfrac{39}{52} = \dfrac{3}{4}

The probability of not drawing a blue card is 3/4, or a 3 in 4 chance.

**b)**

odds

= \dfrac{number \,of\, cards\, that \,are\, not\, blue}{number\, of \,cards \,that \,are\, blue}

= \dfrac{52 \,cards \,-\, 13 \,blue \,cards}{13}

= \dfrac{39}{13} = \dfrac{3}{1}

The odds are 3 to 1 *for* drawing a card that is not blue.