Question

A coin is so biased that appearence of head is twice likely as that of tail.

If a throw is made 6 times, using Binomial distribution, find the probalility that at least two heads will appear.

Answer 1

716/729

Answer 2

Let $P(H)$ be the probability of getting a head and $P(T)$ be the probability of getting a tail.

According to the problem, the appearance of head is twice likely as that of tail, so $P(H) = 2 \cdot P(T)$. We also know that $P(H) + P(T) = 1$. Substituting $P(H) = 2 \cdot P(T)$ into the second equation, we get: $2 \cdot P(T) + P(T) = 1$ $3 \cdot P(T) = 1$ $P(T) = 1/3$ Then, $P(H) = 2 \cdot P(T) = 2 \cdot (1/3) = 2/3$.

Let $p = P(H) = 2/3$ be the probability of success (getting a head) and $q = P(T) = 1/3$ be the probability of failure (getting a tail) in a single throw. The coin is thrown 6 times, so the number of trials is $n = 6$. The number of heads in 6 throws follows a Binomial distribution $B(n, p)$, where $n=6$ and $p=2/3$. Let $X$ be the random variable representing the number of heads in 6 throws. The probability of getting exactly $k$ heads in $n$ throws is given by the Binomial probability formula: $P(X=k) = \binom{n}{k} \cdot p^k \cdot q^{n-k}$

We need to find the probability that at least two heads will appear, which is $P(X \ge 2)$. $P(X \ge 2) = P(X=2) + P(X=3) + P(X=4) + P(X=5) + P(X=6)$. It is easier to calculate the complement probability, which is $P(X < 2) = P(X=0) + P(X=1)$, and then use the formula $P(X \ge 2) = 1 - P(X < 2)$.

Calculate $P(X=0)$: $P(X=0) = \binom{6}{0} \cdot (2/3)^0 \cdot (1/3)^{6-0}$ $P(X=0) = 1 \cdot 1 \cdot (1/3)^6$ $P(X=0) = 1 / 729$

Calculate $P(X=1)$: $P(X=1) = \binom{6}{1} \cdot (2/3)^1 \cdot (1/3)^{6-1}$ $P(X=1) = 6 \cdot (2/3) \cdot (1/3)^5$ $P(X=1) = 6 \cdot (2/3) \cdot (1/243)$ $P(X=1) = (12/3) \cdot (1/243)$ $P(X=1) = 4 \cdot (1/243)$ $P(X=1) = 4/243$ To add $P(X=0)$ and $P(X=1)$, we can express $P(X=1)$ with a denominator of 729: $4/243 = (4 \cdot 3) / (243 \cdot 3) = 12/729$. So, $P(X=1) = 12/729$.

Calculate $P(X < 2)$: $P(X < 2) = P(X=0) + P(X=1)$ $P(X < 2) = 1/729 + 12/729$ $P(X < 2) = 13/729$

Calculate $P(X \ge 2)$: $P(X \ge 2) = 1 - P(X < 2)$ $P(X \ge 2) = 1 - 13/729$ $P(X \ge 2) = 729/729 - 13/729$ $P(X \ge 2) = (729 - 13) / 729$ $P(X \ge 2) = 716 / 729$

The probability that at least two heads will appear is 716/729.

Explanation: The probability of getting a head is $p=2/3$ and a tail is $q=1/3$. For 6 independent throws, the number of heads X follows a Binomial distribution B(6, 2/3). We need $P(X \ge 2)$. This is calculated as $1 - P(X < 2) = 1 - [P(X=0) + P(X=1)]$. $P(X=0) = \binom{6}{0} (2/3)^0 (1/3)^6 = 1/729$. $P(X=1) = \binom{6}{1} (2/3)^1 (1/3)^5 = 6 \cdot (2/3) \cdot (1/243) = 4 \cdot (1/243) = 4/243 = 12/729$. $P(X < 2) = 1/729 + 12/729 = 13/729$. $P(X \ge 2) = 1 - 13/729 = (729 - 13)/729 = 716/729$.