Binomial probability distribution can be applied on such events where only two outcomes are possible. If n repeated trials are done on a binomial experiment then for a certain even p to occur r times, binomial distribution is used. The formula for binomial probability distribution is: $P(X)=^{n}_{x}\textrm{C}\ p^{x}q^{n-x}$
Example 1:

We have a coin and after toss we can have only two outcomes, Head or Tail. Suppose we toss the coin 10 times. If we want all of the ten outcomes to be head then what will be the probability for this?


In one toss for head to come as outcome the probability is $\frac{1}{2}$. Hence, for ten tosses the probability will be $(\frac{1}{2})^{10}$

Hence, Probability, $P$ = $\frac{1}{1024}$

Example 2:

If a coin is tossed ten times then what will be the probability for getting 7 heads and 3 tails in 10 tosses?


For, 7 heads it will be $(\frac{1}{2})^{7}$ and for 3  tails probability will be $(\frac{1}{2})^{3}$. But still one more point is needed to get the complete

solution. Out of 10 tosses, which are those seven who will give head. The number of possible events will be $_{10}^{7}\textrm{C}$. Hence, the

probability to get 7 heads out of 10 tosses of a coin will be,

$P =\ ^{10}_{7}\textrm{C}$ $(\frac{1}{2})^{7} (\frac{1}{2})^{3}$
    = $120\times$ $\frac{1}{128}$ $\times$ $\frac{1}{8}$
    = $\frac{15}{128}$ 

Binomial Probability Word Problems

Problem 1:

In a store, out of all the people who came there thirty percent bought a shirt. If four people came in the store together then find the probability of one of them buying a shirt.


The probability of buying a shirt will be 0.3 and not buying a shirt is given as 1 - 0.3 = 0.7. Now, if one person will buy a shirt out of four, then the probability for this event will be:

$P$ = $^{4}_{1}\textrm{C} (0.3)^{1} (0.7)^{3}$
   = $4\times 0.3\times 0.243$
   = $0.2916$
Problem 2:

In a hospital sixty percent of patients are dying of a disease. If on a certain day, eighth patients got admitted in the hospital for that disease what are the chances of three to survive?


The probability of the patient dying because of the disease is 0.6. So, the probability of the patient being cured is 1 - 0.6 = 0.4. Out of the 8 patients, the probability survival of 3 patients will be,

$P$ = $^{8}_{3}\textrm{C} (0.4)^{3} (0.6)^{5}$
   = $56\times 0.064\times 0.07776$
   = $0.27869184$
Problem 3:

Suppose we are throwing a dice thrice. Find the probability of finding a multiple of 3 in one of the throws.


There are two cases of getting a multiple of 3, that is, 3 and 6. Hence, the probability of getting a multiple of three will be $\frac{2}{6}$ = $\frac{1}{3}$

The probability of not getting a multiple of 3 is $1$ -$\frac{1}{3}$ = $\frac{2}{3}$.

Hence, the probability of getting a multiple of 3 in one out of three events will be,

$P$ = $^{3}_{1}\textrm{C} (\frac{1}{3})^{1} (\frac{2}{3})^{2}$

   = $3 \times$ $\frac{1}{3}$ $\times$ $\frac{2}{9}$

   = $\frac{2}{9}$
Problem 4:

In a restaurant seventy percent of people order for Chinese food and thirty percent for Italian food. A group of three persons enter the restaurant. Find the probability of at least two of them ordering for Italian food.


The probability of ordering Chinese food is 0.7 and the probability of ordering Italian food is 0.3. Now, if at least two of them are ordering Italian food then it implies that either two or three will order Italian food.

Probability for two ordering Italian food,

$P(X = 2)$ = $^{3}_{2}\textrm{C} (0.3)^{2} (0.7)^{1}$

             = $3\times 0.09\times 0.7$
             = $0.189$

Probability for all three ordering Italian food,

$P(X = 3)$ = $_{3}^{3}\textrm{C} (0.3)^{3} (0.7)^{0}$
             = $1\times 0.081\times 1$
             = $0.081$

Hence, the probability for at least two persons ordering Italian food is,

$P = P(X = 2) + P(X = 3)$
     = $0.189 + 0.081 = 0.27$
Problem 5:

In an exam only ten percent students can qualify. If a group of 4 students have appeared, find the probability that at most one student will qualify?


For at most one student to qualify, either 1 student will qualify or none of the 4 will qualify.

Probability for a student to qualify = $0.1$

Probability for a student to disqualify = $0.9$

The probability of qualification of zero students,

$P(X = 0)$ = $^{4}_{0}\textrm{C} (0.1)^{0} (0.9)^{4}$

            = $1\times 1\times 0.6561$

            = $0.6561$

$P(X = 1)$ = $^{4}_{1}\textrm{C} (0.1)^{1} (0.9)^{3}$

             = $4\times 0.1\times 0.729$
             = $0.0729$

Hence, the probability of at the most one student to qualify out of four will be,

$P = P(X = 0) + P(X = 1)$
   = $0.6561 + 0.0729$
   = $0.729$