Six coins are tossed simultaneously. Find the probability of getting

i. 3 heads

ii. no heads

iii. at least one head

Let p be the probability of getting a head in a toss.

If p is the probability of getting a head in a toss, then q is the probability of getting a tail in a toss.

We have p + q = 1

⇒ q = 1 – p

Let X denote a random variable representing the number of heads in 6 tosses of coin. The probability of getting r heads in n tosses of coins is given by

P (X = r) = ⁿC_rp^rq^n-r

We know the values of n, p, and q.

n = 6

We can re-write the binomial distribution as,

…(A)

(i). We need to find the probability of getting 3 heads.

It is given by,

Probability = P (X = 3)

So, putting r = 3 in equation (A). We get

Thus, the probability of getting 3 heads out of tosses of 6 coins is .

(ii). We need to find the probability of getting no head.

It is given by,

Probability = P (X = 0)

So, putting r = 0 in equation (A). We get

Thus the probability of getting 0 heads out of tosses of 6 coins is .

(iii). We need to find the probability of getting at least 1 head.

It is given by,

Probability = P (X = 1) + P (X = 2) + P (X = 3) + P (X = 4) + P (X = 5) + P (X = 6)

Or

Probability = 1 – P (X = 0)

Let us use the shorter formula.

We know from the result of part (ii),

Using this result, let us find out the probability of getting at least one head.

Then,

Thus, the probability of getting at least one head is .