Question

14. Problem 14 If Mike tasses at fair coin four times, what is the probability thut he gets two beads and two tails? Show your work

A) P(X=2)=0.375
B) 0.25P(X=2)=0.25
C)0.5P(X=2)=0.5
D)0.125P(X=2)=0.125

Answer 1

The probability that Mike gets two heads and two tails when tossing a fair coin four times is calculated using combinations and is found to be 0.375. This is done by dividing the 6 favorable arrangements of two heads and two tails by the 16 total possible outcomes from four coin tosses.

Step-by-step explanation:

To find the probability that Mike gets two heads and two tails when tossing a fair coin four times, we can use the concept of combinations and the basic probability principles for a fair coin. Since the coin is fair, the probability of getting either a head (H) or a tail (T) on any single toss is 0.5. There are a total of 24 or 16 possible outcomes when a coin is tossed four times.

Next, we look for the number of ways to arrange two heads and two tails. This is a combination problem, which can be represented as '4 choose 2' because we are choosing 2 heads out of 4 possible positions. The formula for combinations is given by:

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

Where 'n' is the total number of items, 'k' is the number of items to choose, and '!' denotes factorial.

Here, n = 4 and k = 2, so C(4, 2) = 4! / (2!2!) = (4×3×2×1) / ((2×1)(2×1)) = 6.

Therefore, there are 6 ways to get two heads and two tails. The probability P(X=2) is the number of favorable outcomes divided by the total number of outcomes:

P(X=2) = 6 / 16 = 0.375

So, the correct answer is A) P(X=2)=0.375.