Bernoulli trials are such trials which has only two possible outcomes. For example, tossing a coin will give only two outputs, head or tail. Such a trial is known as Bernoulli trial. If probability of one of the event will be p then the probability of the other event will be (1 - p). The bernoulli probability distribution is written for success in r trials out of n trials as,

$P(X = r) = C_{r}^{n}(p)^r(1-p)^{n-r}$

$P(X = r) = C_{r}^{n}(p)^r(1-p)^{n-r}$

If the probability of a bulb being defective is 0.8, then what is the probability of the bulb not being defective.

Solution:

Probability of bulb being defective, p = 0.8

Probability of bulb not being defective, q = 1 - p = 1 - 0.8 = 0.2

10 coins are tossed simultaneously where the probability of getting head for each coin is 0.6. Find the probability of getting 4 heads.

Solution:

Probability of getting head, p = 0.6

Probability of getting head, q = 1 - p = 1 - 0.6 = 0.4

Probability of getting 4 heads out of 10, $P(X = 4) = C_{4}^{10}(0.6)^4(0.4)^6$ = 0.111476736

In an exam, 10 multiple choice questions are asked where only one out of four answers is correct. Find the probability of getting 5 out of 10 questions correct in an answer sheet.

Solution:

Probability of getting an answer correct, p = $\frac{1}{4}$ = 0.25

Probability of getting an answer correct, q = 1 - p = 1 - 0.25 = 0.75

Probability of getting 5 answers correct, P(X = 5) = $C_{5}^{10}(0.25)^5(0.75)^5$ = 0.05839920044

The chances of a team winning a match is 0.7. Find the probability that the team will win at least one match out of three.

Probability of winning a match, p = 0.7

Probability of winning a match, q = 1 - p = 1 - 0.7 = 0.3

Probability of winning zero matches, P(X = 0) = $C_{0}^{3}(0.7)^0(0.3)^3$ = 0.027

Probability of winning at least one match = 1 - 0.027 = 0.973