Calculating Probabilities for Binomial Random Variables

We have seen that for a binomial random variable XX, the binomial distribution gives us the probability of each possible outcome as

P(X=x)=nCxpx(1p)nxP(X = x) = {}^nC_x p^x(1-p)^{n-x}

We can use this formula to calculate probabilities for a wide range of outcomes involving the variable XX.

For example, we may be presented with the following scenario:

An exam consists of 5 multiple choice questions, with 4 choices available for each question. The passing grade for this exam is 4 correct answers. What is the probability that a student will get 4 correct answers on the exam if they guess the answer to every question?

We can model the number of correct answers XX from guessing as a binomial random variable with n=5n=5 and p=14p=\frac14. Another way we can write this is XBin(5,14)X \sim \operatorname{Bin}(5, \frac14)

In this question, we are being asked to find the probability of the score from guessing being exactly equal to 4. We can formally write this as P(X=4)P(X = 4), and calculate this using the formula for the binomial distribution

P(X=4)=5C4(14)4(114)1P(X=4)=514434P(X=4)=3545P(X=4)=151024\begin{aligned} P(X = 4) &= {}^5C_4 \left(\frac14\right)^4\left(1-\frac14\right)^1 \\ P(X = 4) &= 5 \cdot \frac1{4^4} \cdot \frac34 \\ P(X = 4) &= \frac{3\cdot 5}{4^5} \\ P(X = 4) &= \frac{15}{1024} \end{aligned}

We can also use this formula to calculate probabilities of outcomes that can be expressed as an inequality on the variable XX. We will see that sometimes, if we are clever about the way we approach the problem, we can sometimes avoid large amounts of calculation to do so.

For example, we might be asked the following equation about the same scenario

What is the probability that a student will pass the exam if they guess the answer to every question?

In this question, we are being asked to find the probability of passing the exam, which means the score is at least 4. We can formally write this as P(X4)P(X \geq 4), and calculate this using

P(X4)=P(X=4)+P(X=5)P(X \geq 4) = P(X = 4) + P(X = 5)

and calculate the individual probabilities of each outcome to determine the probability of passing the exam.

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