A coin is biased so that the probability it will come up tails is 0.46. The coin is tossed three times.  Consider a success to be tails.  Using the binomial probability formula, what is the probability the exactly two of the three tosses are tails?

To solve this problem, we will use the binomial probability formula, which is:

P(X=k) = C(n, k) * (p^k) * ((1-p)^(n-k))

where:

• P(X=k) is the probability of k successes in n trials
• C(n, k) is the combination of n items taken k at a time
• p is the probability of success on a single trial
• n is the number of trials
• k is the number of successes

In this case, we want to find the probability that exactly 2 out of 3 tosses are tails. So, n=3 (the number of tosses), k=2 (the number of successful tosses we want), and p=0.46 (the probability of getting tails on a single toss).

First, calculate C(n, k). This is the number of combinations of n items taken k at a time. The formula for this is:

C(n, k) = n! / [k!(n-k)!]

where "!" denotes factorial, which is the product of all positive integers up to that number. So,

C(3, 2) = 3! / [2!(3-2)!] = 6 / [2*1] = 3

Next, calculate p^k, which is the probability of success on a single trial raised to the power of the number of successes. So,

p^k = 0.46^2 = 0.2116

Then, calculate (1-p)^(n-k), which is the probability of failure on a single trial raised to the power of the number of failures. So,

(1-p)^(n-k) = (1-0.46)^(3-2) = 0.54^1 = 0.54

Finally, substitute these values back into the binomial probability formula to get:

P(X=2) = C(3, 2) * (0.46^2) * (0.54^1) = 3 * 0.2116 * 0.54 = 0.342

So, the probability that exactly 2 out of 3 tosses are tails is approximately 0.342, or 34.2%.