Question

if a coin is flipped four times and we are interested in the event \two heads and two tails\, the event will have ____ outcomes in it.
options: 8, 6, 16, 2

Explanation:

Step1: Recall combination formula

To find the number of ways to get \( k \) successes (here, heads) in \( n \) trials (flips), we use the combination formula \( \binom{n}{k}=\frac{n!}{k!(n - k)!} \), where \( n = 4 \) (flips) and \( k = 2 \) (heads, so tails \(=4 - 2=2\)).

Step2: Calculate the combination

Substitute \( n = 4 \) and \( k = 2 \) into the formula:
\( \binom{4}{2}=\frac{4!}{2!(4 - 2)!}=\frac{4!}{2!2!} \)
Since \( n!=n\times(n - 1)\times\cdots\times1 \), we have \( 4!=4\times3\times2\times1 = 24 \), \( 2!=2\times1 = 2 \).
So \( \binom{4}{2}=\frac{24}{2\times2}=\frac{24}{4}=6 \).

Answer:

6 (corresponding to the option with "6")