The probability theory is a branch of mathematics that deals with the chances of occurrence of the events. In probability theory, we usually find the answer of the question, "how many number of times an event may happen when certain given conditions are followed ?".

Few basic and famous events in probability are:**1)** Toss of a single or a pair of coin.**2)** Drawing a particular-color marble or a ball from a pile.**3)** Drawing a card (or cards) from the deck of cards.

There are mainly two type of events in probability theory: **Independent **and **Dependent**.

Independent events are those which are not based upon the result of previous event.

**For Example:** the outcome in the toss of a coin are independent of one another. Also, drawing marbles from a bag randomly do not depend upon what we received earlier.

Dependent events are the events whose outcomes are affected from the outcomes of previous one.

**For Example:** If one ball is drawn from a bag and then it is not put back in the bag, then the result of another draw will depend upon the previous result.

On the basis of these events, the probability can be classified in following two types:**a)** Probability With Replacement (Independent events)**b)** Probability Without Replacement (Dependent events)In this chapter, we are going to learn about the definition of probability without replacement, its calculation and examples.

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Usually what we study is the probability with replacement in which the object is drawn in one event is replaced back. It does not impact the size of the sample. On the other hand, the probability without replacement is a type of probability in which the events are not replaced back.

Thus, these events are dependent events. In probability without replacement, the item is removed permanently from the sample. This means that every time we pick an item from a sample, that item is being removed and 1 is subtracted from the sample size for the next pick.

In probability without replacement, the size of the sample is reduced by the number of objects drawn with each event.

**We may have a look at the following diagram:**

In this diagram, there is a sac having 2 blue and 3 red balls with sample size 5. If one red ball is picked without replacement, then there will be 2 blue and 2 red balls remaining making sample size equal to 4. Similarly, if one blue ball is picked without replacement, then there will be 1 blue and 3 red balls remaining making sample size equal to 4 for the next trial. The formula for the probability without replacement is also same as the general formula of probability. According to this formula:

Probability of an event or P(E) = $\frac{Total\ number\ of\ favorable\ outcomes}{Sample\ size\ or\ total\ outcomes}$

Total number of favorable outcomes are probable results which are in favor of asked event.**For example -** In a toss of a coin, number of favorable outcomes for getting a tail or head is 1, since we can obtain at most 1 tail (or 1 head) in a toss of coin.

Total outcomes are the size of sample used in the event.

**For Example:** In a toss of a coin, the total outcomes are 2, since there are only two possibilities, head and tail, in the toss of a coin.

In order to find the probability without replacement, we need to follow the steps mentioned below:

__Step 1:__ Determine the type of probability. If it is without replacement, it will be reflected in the question. But it may or may not be clearly written in the question. One has to wisely decide whether it should be with or without replacement.

__Step 2:__ If it is probability without replacement, then the result of first event is found by the following formula

Probability of an event or P(E) = $\frac{Total\ number\ of\ favorable\ outcomes}{Sample\ size\ or\ total\ outcomes}$

__Step 3:__ For finding the result of second event, the total outcomes or size of sample is reduced by the number of objects taken out in first event; i.e. number of items drawn in first event are subtracted from the sample size. Now, above formula is applied for calculating probability.

__Step 4:__ For successive events, step 3 is repeated. In this way, the probability without replacement is determined.
The examples based on probability without replacement are given below:

**Example 1:** A bag contains 5 pineapple candies and 3 orange candies. If Jack comes and eats one and after some times eats another, then find the probability of both being pineapple candies?

**Solution:** Since Jack eats the candies, it eventually means that this is the problem of probability without replacement.

Probability of first candy to be pinapple one

$P(1)$ = $\frac{5}{8}$

Probability of second candy to be pinapple one

$P(2)$ = $\frac{4}{7}$

Probability of both being pineapple candies = Probability of occurring both events together

= P(1).P(2)

$\frac{5}{8}$ . $\frac{4}{7}$

$\frac{5}{14}$

= $\frac{5}{14}$

Example 2: A man has 6 pairs of black and 8 pairs of white socks. Every morning he picks a pair of socks without replacement and put them on. Determine the probability that he he picks a black pair today and a blue pair tomorrow ?

**Solution :** Give that this is a problem of probability without replacement.

Probability of picking up black pair today

P(Black) = $\frac{6}{14}$ = $\frac{3}{7}$

Probability of picking up blue pair tomorrow

P(Blue) = $\frac{8}{13}$ = $\frac{8}{13}$

Probability of occurring both events together

= P(Black).P(Blue)

= $\frac{3}{7}$ . $\frac{8}{13}$

= $\frac{24}{91}$

**Example 3:** Three cards are chosen at random from a deck of cards without replacement. What is the probability that all of them are picture cards ?

**Solution :** A deck of cards always has 12 picture cards.

Probability of getting picture card in third draw

P(3) = $\frac{10}{50}$ = $\frac{1}{5}$

Probability of happening three events together

= P(1).P(2).P(3)

= $\frac{3}{13}$.$\frac{11}{51}$.$\frac{1}{5}$

= $\frac{11}{3315}$

Thus, these events are dependent events. In probability without replacement, the item is removed permanently from the sample. This means that every time we pick an item from a sample, that item is being removed and 1 is subtracted from the sample size for the next pick.

In probability without replacement, the size of the sample is reduced by the number of objects drawn with each event.

In this diagram, there is a sac having 2 blue and 3 red balls with sample size 5. If one red ball is picked without replacement, then there will be 2 blue and 2 red balls remaining making sample size equal to 4. Similarly, if one blue ball is picked without replacement, then there will be 1 blue and 3 red balls remaining making sample size equal to 4 for the next trial. The formula for the probability without replacement is also same as the general formula of probability. According to this formula:

Probability of an event or P(E) = $\frac{Total\ number\ of\ favorable\ outcomes}{Sample\ size\ or\ total\ outcomes}$

Total number of favorable outcomes are probable results which are in favor of asked event.

Total outcomes are the size of sample used in the event.

In order to find the probability without replacement, we need to follow the steps mentioned below:

Probability of an event or P(E) = $\frac{Total\ number\ of\ favorable\ outcomes}{Sample\ size\ or\ total\ outcomes}$

Probability of first candy to be pinapple one

$P(1)$ = $\frac{5}{8}$

Probability of second candy to be pinapple one

$P(2)$ = $\frac{4}{7}$

Probability of both being pineapple candies = Probability of occurring both events together

= P(1).P(2)

$\frac{5}{8}$ . $\frac{4}{7}$

$\frac{5}{14}$

= $\frac{5}{14}$

Example 2:

Probability of picking up black pair today

P(Black) = $\frac{6}{14}$ = $\frac{3}{7}$

Probability of picking up blue pair tomorrow

P(Blue) = $\frac{8}{13}$ = $\frac{8}{13}$

Probability of occurring both events together

= P(Black).P(Blue)

= $\frac{3}{7}$ . $\frac{8}{13}$

= $\frac{24}{91}$

Probability of getting picture card in first draw

P(1) = $\frac{12}{52}$ = $\frac{3}{13}$

P(1) = $\frac{12}{52}$ = $\frac{3}{13}$

Probability of getting picture card in second draw

P(2) = $\frac{11}{51}$

P(2) = $\frac{11}{51}$

Probability of getting picture card in third draw

P(3) = $\frac{10}{50}$ = $\frac{1}{5}$

Probability of happening three events together

= P(1).P(2).P(3)

= $\frac{3}{13}$.$\frac{11}{51}$.$\frac{1}{5}$

= $\frac{11}{3315}$

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