Question

Nathan is flipping a coin three times.

What is the probability that it will land on tails twice and heads once?


A.


B.


C.


D.

Answer 1

To find the probability of getting tails twice and heads once in three coin flips, we can use the binomial probability formula.

The formula to calculate the probability of getting exactly \( k \) successes (in this case, tails) in \( n \) trials (flips), where the probability of success on each trial is \( p \), is given by:

\[
P(X = k) = \binom{n}{k} p^k (1-p)^{n-k}
\]

For a fair coin, \( p = 0.5 \) (the probability of getting tails), and \( (1-p) = 0.5 \) (the probability of getting heads). Here, \( n = 3 \) (the total number of flips), and \( k = 2 \) (the number of tails we want).

1. Calculate the binomial coefficient \( \binom{3}{2} \):

\[
\binom{3}{2} = \frac{3!}{2!(3-2)!} = \frac{3 \times 2 \times 1}{(2 \times 1)(1)} = 3
\]

2. Now calculate the probability:

\[
P(X = 2) = \binom{3}{2} (0.5)^2 (0.5)^{3-2} = 3 \times (0.5)^2 \times (0.5)^1
\]

\[
= 3 \times (0.5)^2 \times (0.5) = 3 \times (0.25) \times (0.5) = 3 \times 0.125 = 0.375
\]

So, the probability of getting tails twice and heads once in three coin flips is \( 0.375 \).

If you have multiple choice options, it appears that you may want to express the answer in fraction form or percentage. The fraction form of \( 0.375 \) is \( \frac{3}{8} \) and as a percentage, it's \( 37.5\% \).

Assuming choices A, B, C, and D were fractions or percentages, check which one corresponds to \( 0.375 \) or \( \frac{3}{8} \).