To determine the probability that you will get "heads" no more than once out of 3 flips, we need to consider the scenarios in which you get 0 or 1 head. We can break this down into two parts: the probability of getting 0 heads and the probability of getting 1 head.
We will use the binomial distribution formula to find these probabilities. For a binomial distribution with [tex]\( n \)[/tex] trials and probability [tex]\( p \)[/tex] of success on each trial, the probability of getting exactly [tex]\( k \)[/tex] successes is given by:
[tex]\[ P(X = k) = \binom{n}{k} p^k (1-p)^{n-k} \][/tex]
where [tex]\( \binom{n}{k} \)[/tex] is the binomial coefficient, representing the number of ways to choose [tex]\( k \)[/tex] successes out of [tex]\( n \)[/tex] trials, and is calculated as:
[tex]\[ \binom{n}{k} = \frac{n!}{k!(n-k)!} \][/tex]
Given:
- [tex]\( n = 3 \)[/tex] (the number of flips)
- [tex]\( p = 0.5 \)[/tex] (the probability of getting heads in a single flip)
1. Calculate the probability of getting 0 heads ([tex]\( k = 0 \)[/tex]):
[tex]\[ P_0 = \binom{3}{0} (0.5)^0 (1 - 0.5)^{3-0} \][/tex]
[tex]\[ P_0 = \frac{3!}{0!(3-0)!} (0.5)^0 (0.5)^3 \][/tex]
[tex]\[ P_0 = 1 \cdot 1 \cdot (0.5)^3 \][/tex]
[tex]\[ P_0 = 1 \cdot 0.125 \][/tex]
[tex]\[ P_0 = 0.125 \][/tex]
2. Calculate the probability of getting 1 head ([tex]\( k = 1 \)[/tex]):
[tex]\[ P_1 = \binom{3}{1} (0.5)^1 (1 - 0.5)^{3-1} \][/tex]
[tex]\[ P_1 = \frac{3!}{1!(3-1)!} (0.5)^1 (0.5)^2 \][/tex]
[tex]\[ P_1 = 3 \cdot 0.5 \cdot (0.5)^2 \][/tex]
[tex]\[ P_1 = 3 \cdot 0.5 \cdot 0.25 \][/tex]
[tex]\[ P_1 = 3 \cdot 0.125 \][/tex]
[tex]\[ P_1 = 0.375 \][/tex]
Adding the probabilities together:
[tex]\[ P(\text{0 or 1 heads}) = P_0 + P_1 \][/tex]
[tex]\[ P(\text{0 or 1 heads}) = 0.125 + 0.375 \][/tex]
[tex]\[ P(\text{0 or 1 heads}) = 0.5 \][/tex]
Therefore, the probability that you will get "heads" no more than once out of 3 flips is [tex]\( \boxed{0.5} \)[/tex].
