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# Binomial Distributions and Probability

## P(X successes in 'n' trials): P(X=a) = nCa x p^a x q^(n-a)

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Binomial Distributions and Probability

You've just read the following statement in an online article about getting into the college of your choice. It states: The probability of a student getting into his or her college of choice is 65%. What is the probability that a student won't be accepted into his or her selected college? Why do you think your answer is correct?

### Guidance

In Chapter 4, we will learn more about binomial distributions. However, we are now talking about probability distributions, and as such, we should at least see how the problems change for these distributions. We will briefly introduce the concept and its formula here, and then we will get into more detail in Chapter 4.

The randomness of an individual outcome occurs when we take 1 event and repeat it over and over again. One example is if you were to flip a coin multiple times. In order to calculate the probability of this type of event, we need to look at one more formula.

The probability of getting X\begin{align*}X\end{align*} successes in n\begin{align*}n\end{align*} trials is given by:

P(X=a)=nCa×pa×q(na)\begin{align*}P(X = a) = {_n}C_a \times p^a \times q^{(n - a)}\end{align*}

where:

a\begin{align*}a\end{align*} is the number of successes from the trials.

p\begin{align*}p\end{align*} is the probability of the event occurring.

q\begin{align*}q\end{align*} is the probability of the event not occurring.

Now, remember that in previous concepts, you learned about the formula for combinations; nCr\begin{align*}{_n}C_r\end{align*}. The formula is shown below:

nCr=n!r!(nr)!\begin{align*}{_n}C_r = \frac{n!}{r!(n-r)!}\end{align*}

This combination tells us how many ways we can have 'r' successes in 'n' trials.

Interestingly, it was Blaise Pascal (pictured below) with Pierre de Fermat who provided the world with the basics of probability. These 2 mathematicians studied many different theories in mathematics, one of which was odds and probability. To learn more about Pascal, go to http://en.wikipedia.org/wiki/Blaise_Pascal. To learn more about Fermat, go to http://en.wikipedia.org/wiki/Fermat. These 2 mathematicians have contributed greatly to the world of mathematics.

#### Example A

The probability of scoring above 75% on a math test is 40%. What is the probability of scoring below 75%?

P\begin{align*}P\end{align*}(scoring above 75%) =0.40\begin{align*}= 0.40\end{align*}

Therefore, P\begin{align*}P\end{align*}(scoring below 75%) =10.40=0.60\begin{align*}= 1 - 0.40 = 0.60\end{align*}.

Now let’s try a few problems with the binomial distribution formula.

#### Example B

A fair die is rolled 10 times. Let X\begin{align*}X\end{align*} be the number of rolls in which we see a 2.

(a) What is the probability of seeing a 2 in any one of the rolls?

(b) What is the probability of seeing a 2 exactly once in the 10 rolls?

(a) P(X)=16=0.167\begin{align*}P(X) = \frac{1}{6} = 0.167\end{align*}

(b) p=0.167q=10.167=0.833n=10a=1\begin{align*}&p = 0.167\\ &q =1 - 0.167 = 0.833\\ &n = 10\\ &a = 1\end{align*}

P(X=a)P(X=1)P(X=1)P(X=1)P(X=1)=nCa×pa×q(na)=10C1×p1×q(101)=10C1×(0.167)1×(0.833)(101)=10×0.167×0.193=0.322\begin{align*}P(X = a) & = {_n}C_a \times p^a \times q^{(n - a)}\\ P(X = 1) & = {_{10}}C_1 \times p^1 \times q^{(10-1)}\\ P(X = 1) & = {_{10}}C_1 \times (0.167)^1 \times (0.833)^{(10-1)}\\ P(X = 1) & = 10 \times 0.167 \times 0.193\\ P(X = 1) & = 0.322\end{align*}

Therefore, the probability of seeing a 2 exactly once when a die is rolled 10 times is 32.2%.

#### Example C

A fair die is rolled 15 times. Let X\begin{align*}X\end{align*} be the number of rolls in which we see a 2.

(a) What is the probability of seeing a 2 in any one of the rolls?

(b) What is the probability of seeing a 2 exactly twice in the 15 rolls?

(a) P(X)=16=0.167\begin{align*}P(X) = \frac{1}{6} = 0.167\end{align*}

(b) pqna=0.167=10.167=0.833=15=2\begin{align*}p &= 0.167\\ q &= 1 - 0.167 = 0.833\\ n &= 15\\ a &= 2\end{align*}

P(X=a)P(X=2)P(X=2)P(X=2)P(X=2)=nCa×pa×q(na)=15C2×p2×q(152)=15C2×(0.167)2×(0.833)(152)=105×0.0279×0.0930=0.272\begin{align*}P(X = a) & = {_n}C_a \times p^a \times q^{(n - a)}\\ P(X = 2) & = {_{15}}C_2 \times p^2 \times q^{(15-2)}\\ P(X = 2) & = {_{15}}C_2 \times (0.167)^2 \times (0.833)^{(15-2)}\\ P(X = 2) & = 105 \times 0.0279 \times 0.0930\\ P(X = 2) & = 0.272\end{align*}

Therefore, the probability of seeing a 2 exactly twice when a die is rolled 15 times is 27.2%.

It should be noted here that the examples involve binomial experiments. Again, we will be learning more about binomial experiments and distributions in Chapter 4. For now, we can visualize a binomial distribution experiment as one that has a fixed number of trials, with each trial being independent of the others. In other words, rolling a die twice to see if a 2 appears is a binomial experiment, because there is a fixed number of trials (2), and each roll is independent of the others. Also, for binomial experiments, there are only 2 possible outcomes (a successful event and a non-successful event). For our rolling of the die, a successful event is seeing a 2, and a non-successful event is not seeing a 2.

### Vocabulary

Thebinomial distribution, which is a distribution produced by an experiment

1. With 2 possible outcomes; success or failure
2. Where there is a fixed number of trials and successes in X\begin{align*}X\end{align*} (random variable)
3. Each trial is independent of the others
4. The probability of a success in each trial is constant

### Guided Practice

A pair of fair dice is rolled 10 times. Let X\begin{align*}X\end{align*} be the number of rolls in which we see at least one 2.

(a) What is the probability of seeing at least one 2 in any one roll of the pair of dice?

(b) What is the probability that in exactly half of the 10 rolls, we see at least one 2?

If we look at the chart below, we can see the number of times a 2 shows up when rolling 2 dice.

(a) The probability of seeing at least one 2 in any one roll of the pair of dice is:

P(X)=1136=0.306\begin{align*}P(X) = \frac{11}{36} = 0.306\end{align*}

(b) The probability of seeing at least one 2 in exactly 5 of the 10 rolls is calculated as follows:

pqna=0.306=10.306=0.694=10=5\begin{align*}p & = 0.306\\ q & = 1 - 0.306 = 0.694\\ n & = 10\\ a & = 5\end{align*}

P(X=a)P(X=5)P(X=5)P(X=5)P(X=5)=nCa×pa×q(na)=10C5×p5×q(105)=10C5×(0.306)5×(0.694)(105)=252×0.00268×0.161=0.109\begin{align*}P(X = a) & = {_n}C_a \times p^a \times q^{(n - a)}\\ P(X = 5) & = {_{10}}C_5 \times p^5 \times q^{(10-5)}\\ P(X = 5) & = {_{10}}C_5 \times (0.306)^5 \times (0.694)^{(10-5)}\\ P(X = 5) & = 252 \times 0.00268 \times 0.161\\ P(X = 5) & = 0.109\end{align*}

Therefore, the probability of rolling at least one 2 exactly 5 times when 2 dice are rolled 10 times is 10.9%.

### Practice

1. The probability of scoring above 80% on a math test is 20%. What is the probability of scoring below 80%?
2. The probability of getting a job after university is 85%. What is the probability of not getting a job after university?
3. A fair die is rolled 10 times. Let X\begin{align*}X\end{align*}be the number of rolls in which we see a 6.
1. What is the probability of seeing a 6 in any one of the rolls?
2. What is the probability that we will see a 6 exactly once in the 10 rolls?
4. A fair die is rolled 15 times. Let X\begin{align*}X\end{align*}be the number of rolls in which we see a 6.
1. What is the probability of seeing a 6 in any one of the rolls?
2. What is the probability that we will see a 6 exactly once in the 15 rolls?
5. A fair die is rolled 15 times. Let X\begin{align*}X\end{align*}be the number of rolls in which we see a 5.
1. What is the probability of seeing a 5 in any one of the rolls?
2. What is the probability that we will see a 5 exactly 7 times in the 15 rolls?
6. A pair of fair dice is rolled 10 times. Let X\begin{align*}X\end{align*}be the number of rolls in which we see at least one 5.
1. What is the probability of seeing at least one 5 in any one roll of the pair of dice?
2. What is the probability that in exactly half of the 10 rolls, we see at least one 5?
7. A pair of fair dice is rolled 15 times. Let X\begin{align*}X\end{align*}be the number of rolls in which we see at least one 5.
1. What is the probability of seeing at least one 5 in any one roll of the pair of dice?
2. What is the probability that in exactly 8 of the 15 rolls, we see at least one 5?
8. A coin is flipped 25 times. Let X\begin{align*}X\end{align*}be the number of flips in which we see a head.
1. What is the probability of seeing a head in any one flip of the coin?
2. What is the probability that we see a head in exactly 11 of the 25 flips?
9. A coin is flipped 30 times. Let X\begin{align*}X\end{align*}be the number of flips in which we see a head.
1. What is the probability of seeing a head in any one flip of the coin?
2. What is the probability that we see a head in exactly 20 of the 30 flips?
10. A pair of coins is tossed 30 times. Let X\begin{align*}X\end{align*}be the number of tosses in which we see at least one head.
1. What is the probability of seeing at least one head in any one toss of the pair of coins?
2. What is the probability that in exactly 25 of the 30 tosses, we see at least one head?

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