**Question 1: **A fair coin is tossed 6 times and the probability of heads on any toss is noted as 0.3. Calculate

- P(X = 2)

2. P(1 < X $\leq$ 5)

** Solution: **

The Probability mass function of binomial distribution is given by

$ f(x) = P(X=x) = ^{n}C_{x}\ p^{x}(1-p)^{n-x}$ ; x = 0, 1, 2, 3, ...., n

X : Number of heads

Suppose now heads are taken as success then we have n = 6, x = 2 and p = 0.3, q = (1 - p) = (1 - 0.3) = 0.7

Now the Probability mass function for the given problem is

1. P(X = 2) = $\binom{6}{2}(0.3)^{2}(0.7)^{4} = 0.324$

2. Now for P(1 < X $\leq$ 5) = P(x = 2) + P(x = 3) + P(x = 4) + P(x =5)

Consider,

P(X = 2) = $\binom{6}{2}(0.3)^{2}(0.7)^{4} = 0.324$

P(X = 3) = $\binom{6}{3}(0.3)^{3}(0.7)^{3} = 0.185$

P(X = 4) = $\binom{6}{4}(0.3)^{4}(0.7)^{2} = 0.059$

P(X = 5) = $\binom{6}{5}(0.3)^{5}(0.7)^{1} = 0.01$

So, P(1 < X $\leq$ 5) = 0.324 + 0.185 + 0.089 + 0.01

= 0.58

**Question 2: **10 fair coins are tossed simultaneously find the probability of getting

Exactly 7 heads

Atleast 7 heads

Atmost 7 heads

** Solution: **

Random Experiment : Tossing 10 coins simulataneously

X : Number of heads obtained

X $\sim$ B(n, p), X $\sim$ B(10, 0.5), Given n= 10 and p = 0.5 (fair coin)

The probability mass function of a binomial random variable X with parameters n and p is given as:

$ f(x) = P(X = x) = ^{n}C_{x}\ p^{x}(1-p)^{n-x}$ ; x=0,1,2,3,....,n $0\leq p\leq 1$, q = 1 - p

1) Now to calculate P(Getting exactly 7 heads)

$P(x = 7)$ =$\binom{10}{7}$ $\ 0.5^{7}$ $\ 0.5^{3}$ = 0.1171

2) To calculate atleast 7 heads

P(X $\geq$ 7) = P(X = 7) + P(X = 8) +P(X = 9) +P(X = 10)

Consider

P(X = 8) = $\binom{10}{8}$ $\ (0.5)^{8}$ $\ (0.5)^{2}$ $\ =0.04394$

P(X = 9) =$\binom{10}{9}$ $\ (0.5)^{9}$ $\ (0.5)^{1}$ $\ =0.0095$

P(X = 10) =$\binom{10}{10}$ $\ (0.5)^{10}$ $\ (0.5)^{0}$ $\ =0.0009$

Therefore, P(X $\geq$ 7) = P(X = 7) + P(X = 8) + P(X = 9) + P(X = 10)

= 0.1171 + 0.04394 + 0.0095 + 0.0009

=0.1719

3) To calculate atmost 7 heads

P(x $\leq$ 7) = 1 - P(X $\geq$ 7)

= 1 - 0.1719

= 0.9453