Binomial distribution is applied on the experiments which can give only two outcomes. In binomial probability, if probability of success is $p$, then probability of failure is $(1 - p)$. In a binomial experiment, for $x$ number of successes in n trials, where $p$ is the probability of success and $q$ is the probability of failure can be given as: $P(x)$ = $_{x}^{n}\textrm{C}p^{x}q^{n-x}$

To find the probability of number of success less than a given value x, the binomial distribution can be given by,

$P(X = n) = P(X = 0) + P(X = 1) + ... + P(X = n)$

To find the probability of number of success less than a given value x, the binomial distribution can be given by,

$P(X = n) = P(X = 0) + P(X = 1) + ... + P(X = n)$

The probability of winning a match for team $A$ is $0.6$. Find the probability of winning $3$ matches out of $5$.

Probability of winning, $p$ = $0.6$

Probability of losing, $q$ = $0.4$

Probability of winning $3$ matches out of $5,\ P(x = 3)$ = $_{3}^{5}\textrm{C}(0.6)^{3}(0.4)^{2}$

$P(x = 3)$ = $\frac{5!}{3!2!}$ $\times\ 0.216\times\ 0.16 = 0.3456$

Hence, the probability is $0.3456$.

If a committee has $7$ members, find the probability of having more female members than male members given that the probability of having a male or a female member is equal.

The probability of having a female member = $0.5$

The probability of having a male member = $0.5$

To have more female members, the number of females should be greater than or equal to $4$.

$P(X\ \geq\ 4)$ = $P(4) + P(5) + P(6) + P(7)$

= $C_{4}^{7}(0.5)^{4}(0.5)^{3} + C_{5}^{7}(0.5)^{5}(0.5)^{2} + C_{6}^{7}(0.5)^{6}(0.5)^{1} + C_{7}^{7}(0.5)^{7}(0.5)^{0}$

= $(0.5)^{7}\times (C_{4}^{7}+C_{5}^{7}+C_{6}^{7}+C_{7}^{7})= 0.0078125\times 64=0.5$

The probability is $0.5$.

Aren is taking part in four competitions. If the probability of him winning any competition is $0.3$, find the probability of him winning at least one competition.

Probability of winning at least one competition will be the complement of the probability of winning not a single competition.

$P(X = 0)$ = $_{0}^{4}\textrm{C}(0.3)^{0}(0.7)^{4}$

= $1 \times 1 \times 0.2401 = 0.2401$

Chances of winning at least one competition = $1 - P(X = 0) = 1 - 0.2401 = 0.7599$

If a coin is tossed thrice, find the probability of a getting head at least two times.

The probability of getting head at least two times is the sum of probabilities of getting head two times and three times.

$P(X\ \geq\ 2)$ = $P(X = 2) + P(X = 3)$

= $_{2}^{3}\textrm{C}(0.5)^{2}(0.5)^{1} + _{3}^{3}\textrm{C}(0.5)^{3}(0.5)^{0}$

= $3\times 0.125 + 1\times 0.125 = 0.5$

Hence, the needed probability is $0.5$.

If only $5$ percent kids can secure A grade in a paper, find the probability of at most $2$ out of $10$ kids getting A grade in that paper.

Probability of securing grade $A = 0.05$

Probability of at most $2$ kids getting grade $A = P(X = 0) + P(X = 1) + P(X = 2)$

The required probability = $C_{0}^{10}(0.05)^{0}(0.95)^{10} + C_{1}^{10}(0.05)^{1}(0.95)^{9} + C_{2}^{10}(0.05)^{2}(0.95)^{8}$

= $0.59873693924 + 0.31512470486 + 0.07463479852 = 0.98849644262$

Hence, the probability needed is $ 0.98849644262$.