**Introduction to probability questions:**

Probability is the likelihood of the occurrence of an event. The probability of event A is written P(A). Probabilities are always numbers between 0 (impossible) and 1(possible), inclusive. Set of possible outcomes of a particular experiment is called as event. The set of all possible outcomes of an experiment is referred to as sample space. For example a coin is tossed, sample space {head, tail}.

**Solved probability questions**

**Example 1:** A bag contains 25 bulbs, out of 25 bulbs, 8 bulbs are defective, 4 bulbs are chosen at random from this bag. Find the probability that at least two of these is
defective.

**Solution:**

Out of 25 bulbs, 8 bulbs are defective.

17 bulbs are favorable bulbs.

E = event for getting no bulb is defective.

n(E) = ^{17}C_{4}

Out of 17 bulbs we have to choose 4 bulbs randomly, so the number of ways = ^{17}C_{4}.

n(E) = ^{17}C_{4}

n(S) = ^{25}C_{4}

P(E) = ^{17}C_{4}/^{25}C_{4}

= 238/1265

Probability of at least two is defective + probability of two is non defective = 2

P(E) + p(E) = 2

238/1265 + p(E) = 2

P(E) = 2292/1265

**Example 2:** A bag contains 8 red and 6 orange bulbs. 2 bulbs are drawn at random. Find the probability that they are of the same color.

**Solution:**

Let S be the sample space

Number of ways for drawing 2 bulbs out of 8 red and 6 orange bulbs = ^{14}C_{2}

= 91

n(S) = 91

Let E = event of getting both bulbs of the same color

Then

n(E) = number of ways of drawing (2 bulbs out of 8) or (2 balls out of 6)

= ^{8}C_{2} + ^{6}C_{2}

= 28 + 15 = 43

P(E) = n(E)/n(S) = 43/91

Solve these practice questions. These questions are very easy to solve.

**Question 1:** Find the probability of drawing 4 bulbs out of 6 bulbs in bag.

**Answer:** 2/3

**Question 2:** Find the probability of drawing 2 bulbs out of 4 red bulbs and 4 orange bulbs in a bag.

**Answer:** 1/4.