When we find probability without replacement, this implies the events in the probability to be determined are dependent on each other.

When two events are dependent on each other this means that the occurrence of first event affects the occurrence of next event occurring which further affects the probability of the events.

The probability of the first event in line can be easily determined in such cases.

But the probability of the next event in line has to be carefully noticed and then found out as the occurrence of first will affect the occurrence of this event. What usually happens is generally the sample space is changed as without replacement condition deducts the number of sample space by 1. The event may or may not be affected as the cases vary but the sample space differs and thus the probability changes.

To find the combined probability of events we
make use of conditional probability criteria:

The conditional probability of an event with respect to another event is the probability of occurrence of the event after the first event has taken place already.

The conditional probability of an event with respect to another event is the probability of occurrence of the event after the first event has taken place already.

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

Consider a well shuffled deck of 52 cards. Now we draw cards from it one by one without replacing the previous one back.

Consider an event of drawing the first card an ace. This can be determined easily as the deck is complete.

Now if we take second card, then the sample space is now 51 (52 -
1). Let the second event is to draw a red card next, then we have two
cases. The first case is that the first card is black ace and the second
card is a red card and the second case is that the first card is red
ace and then next card is red card.

Here the
probability will be adding these two conditional probabilities to get
the second event probability. If the first card is black then only
sample space is changed and number of red card remain as it is. If the
first card is red ace, then number of red cards is also changed along
with total number of cards.

Thus all these
things make a big difference in finding probability of dependent events
or finding probability without replacement.

Let A be the event of drawing a green candy first.

Then P (A) = $\frac{5}{15}$

Now since we are not replacing back, thus, number of green candies left in bag now is 4 and total number of candies is 14.

Let B be the vent of drawing a green candy again.

Then P (B|A) = $\frac{4}{14}$.

Thus the probability of getting both candies green = P (A & B)

= P (A) * P (B|A)

= $\frac{5}{15}$ * $\frac{4}{14}$

= $\frac{1}{3}$ * $\frac{2}{7}$ = $\frac{2}{21}$

**Example 2:** A jar contains 10 blue balls and 11 red balls. Two balls are drawn without replacement. What is the probability of getting two red balls.

**Solution:** Total number of balls = 10 + 11 = 21

Let P(A) = Probability of getting first red ball and

P(B) = Probability of getting second red ball

Therefore:

P(A) = $\frac{11}{21}$

After first withdraw we are left with 20 balls.

So P(B) = $\frac{10}{20}$

Probability of getting two red balls = P(A)P(B) = $\frac{11}{21}$ $\times$ $\frac{10}{20}$

= $\frac{110}{420}$ = $\frac{11}{42}$

Let P(A) = Probability of getting first red ball and

P(B) = Probability of getting second red ball

Therefore:

P(A) = $\frac{11}{21}$

After first withdraw we are left with 20 balls.

So P(B) = $\frac{10}{20}$

Probability of getting two red balls = P(A)P(B) = $\frac{11}{21}$ $\times$ $\frac{10}{20}$

= $\frac{110}{420}$ = $\frac{11}{42}$