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Conditional Probability

P(B|A) = P(A and B)/ P(A)

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Calculating Conditional Probabilities

Suppose you wanted to calculate the probability of pulling the King of Hearts, then the Jack of Diamonds, and then any of the four Aces, from a standard deck of 52 cards, in that order, and without replacing any cards between pulls. Would the probability be significantly different than if you put the cards back after drawing each time?

Credit: OpenClips
Source: http://pixabay.com/en/card-deck-deck-cards-playing-cards-161536/?oq=card%20deck
License: CC BY-NC 3.0

In this lesson we will discuss conditional probabilities that are different for each trial. We’ll return to this question after the lesson.

Conditional Probabilities 

In a previous lesson, we discussed compound probabilities and reviewed some situations involving the probability of multiple occurrences of the same event in a row. A standard example would be the probability of throwing a fair coin three times and getting three heads. In this lesson, we will be introducing a slightly more complex situation, where the coin may or may not be fair.

A new concept we will be introducing in this lesson is the “given that” concept. The idea is that we sometimes need to calculate a probability with a specific condition, for example:

The probability of rolling a “2” on a standard die is \begin{align*}\frac{1}{6}\end{align*}. What is the probability of rolling a “2”, given that I know already that I have rolled an even number? As described in the video above, this is a conditional probability, and we notate it this way: \begin{align*}P(2|even)\end{align*}, which is read as “The probability of rolling a “2” given that we roll an even number”.

The difference in calculations is:

\begin{align*}P(2)&=\frac{1}{6}( \text{on number on the 6-sided die is a 2}) \\ P(2|even)&=\frac{1}{3} ( \text{one number out of the three even numbers is a 2}) \\ \end{align*}

To calculate a “given that” type of problem, we use the conditional probability formula:

\begin{align*}P(A|B)=\frac{P(A \cap B)}{P (B)}\end{align*}

This is read as: “The probability that \begin{align*}A\end{align*} will occur, given that \begin{align*}B\end{align*} will occur (or has occurred), is equal to the intersection of probabilities \begin{align*}A\end{align*} and \begin{align*}B\end{align*} divided by the probability of \begin{align*}B\end{align*} alone.

We have practiced the use of the addition rule and the multiplication rule for calculating probabilities, here we will also be using those again, but this time we will need to combine them for some of the problems.

For review:

Multiplication Rule for independent events: \begin{align*}P(A \ then \ B)= P(A)\times P(B)\end{align*}

Addition Rule for mutually exclusive events: \begin{align*}P(A \ or \ B)=P(A)+P(B) \end{align*}

Addition Rule for mutually inclusive events: \begin{align*}P(A \ or \ B)=P(A)+P(B)-P(A \ and \ B)\end{align*}

Calculating Probability 

1. What is the probability that you have pulled the Jack of Hearts from a standard deck, given that you know you have pulled a face card?

Let’s solve this using the conditional probability formula first (A), then check by looking at the question another way (B):

A. The problem asks us to calculate the probability of a card being the Jack of Hearts, given that the card is a face card: \begin{align*}P(Jack \ of \ Hearts|face \ card)\end{align*}. Apply the conditional probability formula: \begin{align*}P(A|B)=\frac{P(A \cap B)}{P (B)}\end{align*}. Putting in the information from the problem gives us:

\begin{align*}P(Jack \ of \ Hearts|face \ card)&=\frac{P(Jack \ of \ Hearts \cap face \ card)}{P(face \ card)} \\ P(Jack \ of \ Hearts|face \ card)&= \frac{\left(\frac{1}{52}\right)}{\frac{12}{52}} \\ P(Jack \ of \ Hearts|face \ card)&=\frac{1}{52} \times \frac{52}{12}=\frac{1}{12} \\ P(Jack \ of \ Hearts|face \ card)&=\frac{1}{12} \ or \ 8.33 \%\end{align*}

B. The other way to view this is that we are looking for the probability of pulling the Jack of Hearts from the sample space including only face cards, which means we are looking for one specific card from a set including only 12 cards:

\begin{align*}P(Jack \ of \ Hearts)&=\frac{1 \ Jack \ of \ Hearts}{12 \ face \ cards}=\frac{1}{12} \\ P(Jack \ of \ Hearts)&=\frac{1}{12} \ or \ 8.33 \%\end{align*}

2. We calculate 8.33% both ways, looks like we got it!

What is the probability that you could roll a standard die and get a 6, then grab a deck of cards and pull the King of Clubs, keep it, and then pull the Jack of Hearts?

Credit: Images Money
Source: https://www.flickr.com/photos/59937401@N07/5857256935
License: CC BY-NC 3.0

This one looks rather complex, but it can be seen as just three individual probabilities:

  1. \begin{align*}P(roll \ 6)=\frac{1 \ side \ with \ a \ 6}{6 \ sides}=\frac{1}{6}\end{align*}
  2. \begin{align*}P(King)=\frac{1 \ King \ of \ Clubs}{52 \ cards}=\frac{1}{52}\end{align*}
  3. \begin{align*}P(Jack)=\frac{1 \ Jack \ of \ Hearts}{51 \ cards \ left \ after \ first \ pull}=\frac{1}{51} \end{align*}

The overall probability can, and should, be calculated with the multiplication rule, since the 2nd and 3rd are dependent:

\begin{align*}P(roll \ 6 | King | Jack)&=P(roll \ 6)\times P(King)\times P(Jack) \\ P(roll \ 6 \ then \ pull \ King \ then \ pull \ Jack)&=\frac{1}{6}\times \frac{1}{52}\times \frac{1}{51}=\frac{1}{15912} \ OR \ .167\times .019\times .020=.000063\end{align*}

3. You reach into a bag containing 6 coins, 4 are ‘fair’ coins (they have an equal chance of heads or tails), and 2 are ‘unfair’ coins (they have only a 35% chance of tails). If you randomly grab a coin from the bag and flip it 3 times, what is the probability of getting 3 heads?

We actually have two different situations here:

  1. We flip a ‘fair’ coin 3 times and get 3 heads
  2. We flip an ‘unfair’ coin and get 3 heads

Since there are 4 fair coins, and 2 unfair coins, we can say the probability of: \begin{align*}P(choose \ fair)=\frac{4}{6} \ or \ \frac{2}{3}\end{align*} and \begin{align*}P(choose \ unfair)=\frac{2}{6}=\frac{1}{3}\end{align*}.

Note that \begin{align*}P(choose \ unfair)\end{align*} would be the same thing as \begin{align*}P(choose fair)^\prime\end{align*}, (see the apostrophe?) which is the complement of \begin{align*}P(choose \ fair)\end{align*}. In other words: the probability of choosing an unfair coin is 100% minus the probability of choosing a fair coin.

Let’s calculate the probabilities of flipping each 3 times using the multiplication rule:

  • The fair coin has a .5 chance of heads each flip: \begin{align*}P(fair \ 3 \ heads)=.5\times .5\times .5=.125\end{align*}
  • The unfair coin has a .65 chance: \begin{align*}P(unfair \ 3 \ heads)=.65\times .65\times .65= .275\end{align*}

So now we can put them together to find the overall probability (the union) by applying the addition rule:

\begin{align*} P(3 \ heads \ either \ coin)&=.6 \overline{6}\times P(fair \ 3 \ heads)+.3 \overline{3}\times P(unfair \ 3 \ heads) \\ P(3 \ heads \ either \ coin)&=.6 \overline{6}\times .125+.3 \overline{3} \times .275 \\ P(3 \ heads \ either \ coin)&=.083+.092=.175\end{align*}

The probability that we can randomly grab a coin from the bag and flip three heads in a row with it is 17.5%

Earlier Problem Revisited

Suppose you wanted to calculate the probability of pulling the King of Hearts, then the Jack of Diamonds, and then any of the four Aces, from a standard deck of 52 cards, in that order and without putting any back. Would the probability be significantly different than if you put the cards back after drawing each time?

The probability would be different, but perhaps less different than you might think, at least as a percentage. Let’s look at the two cases, with \begin{align*}P(A)\end{align*} representing the probability with each choice coming out of a full deck of 52 cards, and \begin{align*}P(B)\end{align*} representing the probability when the deck gets smaller each pull:

\begin{align*}P(A)&=\frac{1 \ King \ of \ Hearts}{52 \ cards}\times \frac{1 \ Jack \ of \ Diamonds}{52 \ cards}\times \frac{4 \ aces}{52 \ cards}=\frac{1}{52}\times \frac{1}{52}\times \frac{1}{13}=\frac{1}{35152} \\ P(B)&=\frac{1 \ King \ of \ Hearts}{52 \ cards}\times\frac{1 \ Jack \ of \ Diamonds}{51 \ Cards}\times\frac{4 \ aces}{50 \ cards}=\frac{1}{52}\times\frac{1}{51}\times\frac{4}{50}=\frac{4}{132600}=\frac{1}{33150}\end{align*}

The difference in probability is approximately \begin{align*}\frac{1}{2000}\end{align*} or \begin{align*}\frac{5}{100}\end{align*} of 1%, pretty small difference!


Example 1

What would be the probability of the coin landing heads on your first flip>

To calculate the probability of flipping heads, we need to calculate the union of 50% of the probability of flipping heads on each coin. (Why 50% of each probability? There are two coins, so the chance that you will pull either one is 50%)

\begin{align*}P(heads|either \ coin)&=50 \%\times P(heads|fair \ coin)+50 \%\times P(heads|unfair \ coin) \\ P(heads|either \ coin)&=50 \%(50 \%)+50 \%(75 \%) \\ P(heads|either \ coin)&=25 \%+37.5 \% \\ P(heads|either \ coin)&=62.5 \% \end{align*}

Example 2

What would be the probability of flipping tails four times in a row?

To calculate the probability of flipping four tails in a row, we calculate the union of 50% of the probability of flipping four tails in a row with each coin, much like in question 1.

\begin{align*} P(4 \ tails|either \ coin)&=50 \%\times P(4 \ tails|unfair \ coin)+50 \%\times P(4 \ tails|fair \ coin) \\ P(4 \ tails|either \ coin)&=50 \%(25 \%\times 25 \%\times 25 \%\times 25 \%)+50 \%(50 \%\times 50 \% \times 50 \%\times 50 \%) \\ P(4 \ tails|either \ coin)&=0.1953 \%+3.125 \% \\ P(4 \ tails|either \ coin)&=3.32\%\end{align*}

Example 3

What would be the probability of flipping heads five times in a row?

Just like question 2, only this time the probability will end up greater, since the unfair coin has a large chance of heads:

\begin{align*}P(4 \ heads|either \ coin)&=50 \% \times P(4 \ heads|unfair \ coin)+50 \% \times P(4 \ heads|fair \ coin) \\ P(4 \ heads|either \ coin)&=.50\times(.75\times .75\times.75\times .75)^4+.50\times (.25\times .25\times .25\times .25)^4 \\ P(4 \ heads|either \ coin)&=.1582+.0020 \\ P(4 \ heads|either \ coin)&=16.02\% \end{align*}

Example 4

Assume you are using a limited portion of cards that only includes face cards (no number cards). Assume also that you each pull a card, you keep it until the end of the experiment. What would be the probability of pulling three kings in a row?

The key here is to note that you do not replace the card between pulls. That means that the probability changes with each trial. Let’s look at the situation for each trial individuality:

  • \begin{align*}(\text{T}1)\end{align*}: Since we are only dealing with face cards, our first trial will have 12 possible outcomes, four for each of three face cards. Four of the outcomes are favorable, since there are four kings.
  • \begin{align*}(\text{T}2)\end{align*} Our second trial will have only 11 outcomes, since we are keeping the first card. There are only three favorable outcomes this time, since we “used up” a king if \begin{align*}(\text{T}1)\end{align*} was favorable.
  • The third pull \begin{align*}(\text{T}3)\end{align*} only has 10 outcomes, since we will already have the other two cards. Two of the outcomes are favorable, since there would be only two kings left.

\begin{align*}P(three \ kings|face \ card)&=P(T1)\times P(T2)\times P(T3) \\ P(three \ kings|face \ card)&=\frac{4 \ kings}{12 \ face \ cards}\times \frac{3 \ kings}{11 \ face \ cards}\times \frac{2 \ kings}{10 \ face \ cards} \\ P(three \ kings|face \ card)&=.3 \overline{3}\times .27 \overline{27}\times .2 \\ P(three \ kings|face \ card)&=0.0182 \ or \ 1.82\%\end{align*}

Example 5

What would be the probability of rolling a 5, given that you know you rolled an odd number? 

This is a ‘given that’ problem, so we can use the conditional probability formula:

\begin{align*}P(roll \ 5 | roll \ odd)&=\frac{P(roll \ 5) \cap P(roll \ odd)}{P(roll \ odd)} \\ P(roll \ 5 | roll \ odd)&=\frac{\frac{1}{6}}{\frac{1}{2}} \\ P(roll \ 5 | roll \ odd)&=\frac{2}{6} \ or \ \frac{1}{3} \ or \ 33.33\%\end{align*}


  1. What is the probability that you roll two standard dice, and get 4’s on both, given that you know that you have already rolled a 4 on one of them?
  2. Assuming you are using a standard deck, what is the probability of drawing two cards in a row, without replacement, that are the same suit?
  3. What is the probability that a single roll of two standard dice will result in a sum greater than 8, given that one of the dice is a 6?
  4. Assuming a standard deck, what is the probability of drawing 3 queens in a row, given that the first card is a queen?
  5. There are 130 students in your class, 50 have laptops, and 80 have tablets. 20 of those students have both a laptop and a tablet. What is the probability that a randomly chosen student has a tablet, given that she has a laptop?
  6. Thirty percent of your friends like both Twilight and The Hobbit, and half of your friends like The Hobbit. What percentage of your friends who like the Hobbit also like Twilight?
  7. Assume you pull and keep two candies from a jar containing sweet candies and sour candies. If the probability of selecting one sour candy and one sweet candy is 39%, and the probability of selecting a sweet candy first is 52%, what is the probability that you will pull a sour candy on your second pull, given that you pulled a sweet candy on your first pull?
  8. The probability that a student has called in sick and that it is Monday is 12%. The probability that it is Monday and not another day of the school week is 20% (there are only five days in the school week). What is the probability that a student has called in sick, given that it is Monday?
  9. A neighborhood wanted to improve its parks so it surveyed kids to find out whether or not they rode bikes or skateboards. Out of 2300 children in the neighborhood that ride something, 1800 rode bikes, and 500 rode skateboards, while 200 of those ride both a bike and skateboard. What is the probability that a student rides a skateboard, given that he or she rides a bike?
  10. A movie theatre is curious about how many of its patrons buy food, how many buy a drink, and how many buy both. They track 300 people through the concessions stand one evening, out of the 300, 78 buy food only, 113 buy a drink only and the remainder buy both. What is the probability that a patron buys a drink if they have already bought food?
  11. A sporting goods store want to know if it would be wise to place sports socks right next to the athletic shoes. First they keep the socks and shoes in separate areas of the store. They track purchases for one day, Saturday, their busiest day. There were a total of 147 people who bought socks, shoes, or both in one given day. Of those 45 bought only socks, 72 bought only shoes and the remainder bought both. What is the probability that a person bought shoes, if they purchased socks?
  12. The following week they put socks right next to the shoes to see how it would affect Saturday sales. The results were as follows; a total of 163 people bought socks, shoes, or both. Of those 52 bought only socks, 76 bought only shoes and the remainder bought both. What is the probability that a person bought socks, if they purchased shoes?
  13. A florist wanted to know how many roses and daisies to order for the upcoming valentines rush. She used last year’s statistics to determine how many to buy. Last year she sold 52 arrangements with roses only, 15 arrangements with daises only, and 36 arrangements with a mixture of roses and daises. What is the probability that an arrangement has at least one daisy, given that it has at least one rose?

Review (Answers)

To view the Review answers, open this PDF file and look for section 6.6. 

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conditional probability The probability of a particular dependent event  given the outcome of the event on which it occurs.
conditional probability formula The conditional probability formula is P(A/B) = P(AUB)/P(B)
Dependent Events In probability situations, dependent events are events where one outcome impacts the probability of the other.
Favorable Outcome A favorable outcome is the outcome that you are looking for in an experiment.
Independent Events Two events are independent if the occurrence of one event does not impact the probability of the other event.
Multiplication Rule States that for 2 events (A and B), the probability of A and B is given by: P(A and B) = P(A) x P(B).
Mutually Exclusive Events Mutually exclusive events have no common outcomes.
Sample Space In a probability experiment, the sample space is the set of all the possible outcomes of the experiment.

Image Attributions

  1. [1]^ Credit: OpenClips; Source: http://pixabay.com/en/card-deck-deck-cards-playing-cards-161536/?oq=card%20deck; License: CC BY-NC 3.0
  2. [2]^ Credit: Images Money; Source: https://www.flickr.com/photos/59937401@N07/5857256935; License: CC BY-NC 3.0

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