Question

Find probability of getting atleast 3 heads when 5 coins are tossed

Answer 1

1/2

Answer 2

To find the probability of getting at least 3 heads when 5 coins are tossed, we can use the binomial probability formula.

Let:

• $n$ = number of trials (coin tosses) = 5
• $p$ = probability of getting a head (success) = 1/2
• $q$ = probability of getting a tail (failure) = 1 - p = 1/2

The probability of getting exactly $r$ heads in $n$ tosses is given by: $P(X=r) = {}^nC_r * p^r * q^{n-r}$

We need to find the probability of getting at least 3 heads, which means the number of heads can be 3, 4, or 5. So, $P(\text{at least 3 heads}) = P(X=3) + P(X=4) + P(X=5)$

1. Probability of getting exactly 3 heads ($P(X=3)$): $P(X=3) = {}^5C_3 * (1/2)^3 * (1/2)^{(5-3)}$ $P(X=3) = {}^5C_3 * (1/2)^3 * (1/2)^2$ $P(X=3) = {}^5C_3 * (1/2)^5$ ${}^5C_3 = \dfrac{5!}{3!(5-3)!} = \dfrac{5!}{3!2!} = \dfrac{5 \times 4}{2 \times 1} = 10$ $P(X=3) = 10 * (1/32) = 10/32$

2. Probability of getting exactly 4 heads ($P(X=4)$): $P(X=4) = {}^5C_4 * (1/2)^4 * (1/2)^{(5-4)}$ $P(X=4) = {}^5C_4 * (1/2)^4 * (1/2)^1$ $P(X=4) = {}^5C_4 * (1/2)^5$ ${}^5C_4 = \dfrac{5!}{4!(5-4)!} = \dfrac{5!}{4!1!} = 5$ $P(X=4) = 5 * (1/32) = 5/32$

3. Probability of getting exactly 5 heads ($P(X=5)$): $P(X=5) = {}^5C_5 * (1/2)^5 * (1/2)^{(5-5)}$ $P(X=5) = {}^5C_5 * (1/2)^5 * (1/2)^0$ $P(X=5) = {}^5C_5 * (1/2)^5$ ${}^5C_5 = \dfrac{5!}{5!(5-5)!} = \dfrac{5!}{5!0!} = 1$ $P(X=5) = 1 * (1/32) = 1/32$

Now, sum these probabilities: $P(\text{at least 3 heads}) = P(X=3) + P(X=4) + P(X=5)$ $P(\text{at least 3 heads}) = 10/32 + 5/32 + 1/32$ $P(\text{at least 3 heads}) = (10 + 5 + 1) / 32$ $P(\text{at least 3 heads}) = 16/32$ $P(\text{at least 3 heads}) = 1/2$