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.

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:

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 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)

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

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

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

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