Probability is the chance of the outcome of an event of a particular experiment. An event is a one or more possible outcomes of an experiment. Probabilities are occurs always numbers between 0(impossible) and 1(possible). The set of all possible outcomes of a particular experiment is called as sample space. For example, tossing a coin, the possible outcomes are {head, tail}. So, probability of getting a head is `1/2`.

**Example 1: **If throwing two coins simultaneously, what is the probability of getting exactly one head?

**Solution:**

Let S be the sample space, n(S) = {HH, HT, TH, TT} = 4

A be the event of getting exactly one head, n(A) = 2

P(A) = `(n(A))/(n(S))` = `2/4` = `1/2`

Therefore probability of getting exactly one head is `1/2` .

**Example 2:** There are 3 apples and 6 bananas in a bag. If two fruits are drawn at random. What is the probability of getting exactly one apple?

**Solution:**

Let S be the sample space S = 3 + 6 = 9

A be the event of getting exactly one apple.

n(S) = ^{9}C_{2} = `(9!)/(2!xx7!)` = `(9xx8)/(2xx1)` = 36

Here, two fruits are drawn at random, so it may be (apple and apple) or (apple and banana) or (banana and apple) or (banana and banana).

n(A) = ^{3}C_{1} x ^{6}C_{1 }+ ^{6}C_{1} x ^{3}C_{1}

= 3 x 6 + 6 x 3 = 18 + 18 = 36

P(A) = `(n(A))/(n(S))` = `36/36`= 1

Let S be the sample space, n(S) = {HHH, HHT, HTH, HTT, THH, THT, TTH, TTT} = 8

A be the event of getting exactly one head, n(A) =3

B be the event of getting exactly one tail, n(B) = 3

P(A) = `(n(A))/(n(S))` = `3/8`

P(B) = `(n(B))/(n(S))` = `3/8`

P(A or B) = P(A) + P(B) = `3/8`+ `3/8` = `3/4`

Therefore probability of getting exactly one head or tail is `3/4`.

**Problem 1: **If throwing two dice simultaneously, what is the probability of getting sum of exactly 10?

**Problem 2:** There are 4 apples and 7 bananas in a bag. If two fruits are drawn at random. What is the probability of getting
exactly one banana?

**Answer:** **1)** `1/12` **2) **`56/55`