Whenever we say probability with replacement, this implies that the events that are take in the probability are completely independent of each other.
Two events are said to be independent only when the occurrence of one does not at all affect the occurrence of another. In such cases the probability of the events also does not get affected at all by each other. In this section we shall learn about the probabilities of the events with replacement.


The formula for finding the combined probability of events obtained with replacement, we simply find the product of the probabilities of the event directly:

P (A $\cap$ B) = P (A) P (B)

Let us take an example of a well shuffled deck having 52 cards. Assume that we take out cards from the deck one by one that too in such a manner that before the subsequent draw the previous card has been placed back in the deck.

Suppose we draw a 5 first. This probability of this event can be evaluated easily as the deck is complete.
Now let the second card be a black card. Here, we have first placed back the 5 card drawn first, this means the deck is again complete. So if we draw another card from a complete deck it is equivalent of saying that we are making a fresh draw from the deck and thus the probability of the event can be calculated. Thus the combined probability of the two events can be evaluated by multiplying the probabilities of the individual events directly.

If we wouldn’t have replaced back the card in the deck then the probabilities will change and the events would no longer be independent.


We can follow the following steps to determine the probability with replacement:

Step 1: Determine the probabilities of the given events individually.

Step 2: Then the combined probability of all the ‘n’ independent events to be determined is given by:

P $(A_1 \cap A_2 \cap ….. \cap A_n)$ = $P (A_1) * P (A_2) * …. * P (A_n)$

Where $A_1, A_2, …., A_n$ are the ‘n’ number of given independent events,
$P (A_1), P(A_2), …, P (A_n)$ are the probabilities of the ‘n’ events and P $(A_1 \cap A_2 \cap ….. \cap A_n)$ is the probability of simultaneous occurrence of all the events.


Let us few examples on finding probability with replacement mathematically to get a better understanding.

Example 1: Given is a basket of fruits containing 4 oranges, 5 apples and 1 pears. We pick three fruits with replacement from the basket. Find the probability of getting an orange and two apples.

Solution: Here, n (S) = 4 + 5 + 1 = 10

Let A be the event of drawing n orange first.

Then P (A) = $\frac{4}{10}$.

As we have replaced back after every pick, so the events are completely independent.

Let B be the event of drawing an apple.

Then P (B) = $\frac{5}{10}$.

Let C be the event of drawing second apple.

Then P (C) = $\frac{5}{10}$

Thus the probability of getting an orange and two apples with replacement

= P (A & B & C)

= P (A) * P (B) * P (C)

= $\frac{4}{10}$ * $\frac{5}{10}$ * $\frac{5}{10}$

= $\frac{2}{5}$ * $\frac{1}{2}$* $\frac{1}{2}$

= $\frac{1}{2}$.

Example 2: A basket contains 3 apples and 7 oranges. A fruit is drawn and put back in the basket. This process is repeated 3 times. What is the probability that selected three fruits are orange?

Solution: A basket contains 3 apples and 7 oranges.

Total number of fruits = 3 + 7 = 10

Since fruit was put back in the basket after each drawn, clearly this is the case of experiment "with replacement".

The probability for each drawn = $\frac{7}{10}$

i.e. P(orange) = $\frac{7}{10}$

Therefore probability of 3 oranges = $\frac{7}{10}$ $\times$ $\frac{7}{10}$ $\times$ $\frac{7}{10}$

or P(orange $\cap$ orange $\cap$ orange) = $\frac{7}{10}$ $\times$ $\frac{7}{10}$ $\times$ $\frac{7}{10}$

= $\frac{343}{1000}$

= 0.343