A coin is biased so that the probability of obtaining a head is 2/3. The coin is tossed four times. Find the probability of obtaining exactly two heads.

this is a binary expansion

(h + t)⁴

the term you want is...6 h² t²

6 * (2/3)² * (1/3)² = 24/81 = 8/27

To find the probability of obtaining exactly two heads when a biased coin is tossed four times, we can use the binomial probability formula.

The formula for finding the probability of getting exactly k successes in n trials, when the probability of success in a single trial is p, is given by:

P(k) = (nCk) * p^k * (1-p)^(n-k)

Where:
P(k) is the probability of exactly k successes
n is the total number of trials
k is the number of successful outcomes
p is the probability of success in a single trial
(1-p) is the probability of failure in a single trial
nCk is the binomial coefficient, which represents the number of ways to choose k successes from n trials.

In this case, the probability of obtaining a head is p = 2/3, and we want to find the probability of exactly two heads (k = 2) when the coin is tossed four times (n = 4).

Using the binomial probability formula:

P(2) = (4C2) * (2/3)^2 * (1-2/3)^(4-2)

The binomial coefficient (4C2) represents the number of ways to choose 2 successes from 4 trials, and it can be calculated as:

(4C2) = 4! / (2! * (4-2)!)

Simplifying further:

(4C2) = 4! / (2! * 2!)
= (4 * 3 * 2!) / (2! * 2!)
= (4 * 3) / 2
= 6

Now we can substitute the values into the formula:

P(2) = 6 * (2/3)^2 * (1-2/3)^(4-2)
= 6 * (4/9) * (1/3)^2
= 6 * (4/9) * (1/9)
= 24/27
= 8/9

Therefore, the probability of obtaining exactly two heads when a biased coin is tossed four times is 8/9.