How to Solve Conditional and Binomial Probabilities?

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A step-by-step guide to conditional and binomial probabilities

The conditional probability of an event is the probability that the event will occur, given that event A has already occurred. This probability can be written as \(P(B|A)\). The formula for conditional probability is:

\(\color{blue}{P\left(B│A\right)=\:\frac{P\:\left(A\:and\:B\right)}{P\left (A\right)}}\)

Where \(P\left(A\right)\ne 0\), which can be written as follow:

\(\color{blue}{P\left(B│A\right)=\:\frac{P\left(A∩B\right)}{P\left(A\right)}}\)

The binomial probability is called the probability of exactly x successes on n repeated trials in an experiment that has two possible outcomes, (where \(0\le x\le n\), \(p\) is the probability of success, and \(1-p\) is the probability of failure.) and is obtained by the following formula:

\(\color{blue}{P=\:_nC_x\:\:p^x\:\left(1-p\right)^{n-x}}\)

Conditional and Binomial Probabilities – Example 1:

A bag contains \(3\) blue, \(5\) red, and \(10\) white marbles. Two marbles are chosen at random. Given that the first marble is red, find the probability that the second marble is blue. (Assume that the first marble has not been replaced.)

Solution: Use the conditional probability formula to solve this problem: \(P\left(B│A\right)=\:\frac{P\:\left(A\:and\:B\right)}{P\left (A\right)}\)

The probability of selecting red marble: \(P (red)=\frac{5}{3+5+10}=\frac{5}{18}\)

The probability of selecting blue marble: \(P (blue)=\frac{3}{3+4+10}=\frac{3}{17}\)

The probability of selecting red and blue marble: \(P (red and blue)=\frac {5}{18}×\frac{3}{17}=\frac{5}{102}\)

Therefore,\(\:P\left(blue│red\right)\)\(=\frac{P\:\left(red\:and\:blue\right)}{P\:\left(red\right)}=\frac{\frac{5}{102}}{\frac{5}{18}}=\frac{3}{17}\)

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Frequently Asked Questions

What is conditional probability?

Conditional probability \(P(A \mid B)\) is the probability of event \(A\) happening given that event \(B\) has happened. The formula is \[P(A \mid B) = \frac{P(A \text{ and } B)}{P(B)},\] valid as long as \(P(B) > 0\). It’s a way of updating your probability for \(A\) once you learn \(B\) occurred.

Walk me through a conditional probability example

A bag has 5 red and 3 blue marbles. Draw two without replacement. \(P(\text{second blue} \mid \text{first red})\)? After drawing a red, the bag has 4 red and 3 blue (7 total). So \(P(\text{second blue} \mid \text{first red}) = 3/7\). The condition (first red) changed the sample space for the second draw.

What’s the multiplication rule for probability?

\(P(A \text{ and } B) = P(A) \cdot P(B \mid A) = P(B) \cdot P(A \mid B)\). If \(A\) and \(B\) are independent, this simplifies to \(P(A) \cdot P(B)\). For independent events, conditioning on one doesn’t change the other’s probability.

When are events independent?

Events \(A\) and \(B\) are independent if knowing one happened doesn’t change the probability of the other: \(P(A \mid B) = P(A)\). Equivalently, \(P(A \text{ and } B) = P(A) \cdot P(B)\). Coin flips are independent. Drawing cards without replacement is NOT independent – the second draw’s odds depend on the first.

What is binomial probability?

Binomial probability is the probability of getting exactly \(k\) successes in \(n\) independent trials, each with the same success probability \(p\). The formula is \[P(X = k) = \binom{n}{k} p^k (1-p)^{n-k}.\] The four requirements: fixed \(n\), two outcomes per trial, constant \(p\), independent trials.

Walk me through a binomial example

A free-throw shooter makes 80% of shots. She takes 10 shots. Find \(P(\text{exactly 8 made})\). \(n = 10\), \(p = 0.8\), \(k = 8\). \(P = \binom{10}{8} (0.8)^8 (0.2)^2 = 45 \cdot 0.16777 \cdot 0.04 \approx 0.302\). So about a 30% chance of exactly 8 makes – the most likely individual outcome.

How do I find expected value of a binomial?

\(E[X] = np\). For the free-throw example, \(E[X] = 10 \cdot 0.8 = 8\) makes on average. For \(n = 60\) die rolls counting 6s, \(E[X] = 60 \cdot 1/6 = 10\). Variance: \(\text{Var}(X) = np(1-p)\). For free throws: \(\text{Var} = 10 \cdot 0.8 \cdot 0.2 = 1.6\), so \(\sigma \approx 1.26\) makes.

How do I compute “at least k” with a binomial?

Sum the individual probabilities, or use the complement when easier. \(P(X \geq 1) = 1 – P(X = 0)\). \(P(X \geq 2) = 1 – P(X = 0) – P(X = 1)\). For small numbers of values to subtract, the complement is fastest. For “at least most,” sometimes summing forward is easier.

What is Bayes’ theorem?

Bayes’ theorem rewrites a conditional probability when you know it the other way around: \[P(A \mid B) = \frac{P(B \mid A) P(A)}{P(B)}.\] Classic medical-testing example: a disease has 1% prevalence and a test that’s 95% sensitive/specific. Even with a positive test, the probability of actually having the disease can be surprisingly low – Bayes’ theorem gives the correct answer.

Where do conditional and binomial probabilities show up on tests?

AP Statistics, college intro stats, the GRE, the actuarial exams, SAT (lightly), and pre-calc/discrete-math courses. Common question types: compute conditional probability from a two-way table, check whether two events are independent, apply the binomial formula, find the expected value of a binomial, or use Bayes’ theorem to update probabilities.

Related EffortlessMath Lessons

If a topic on this page feels rusty, these short lessons go deeper:

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