Lays a game at a fundraiser for which there is a probability of 0.2 that she wins. She plans to play this 5 times.
[tex]P(X=k)=\binom{n}{k} p^k(1-p)^{n-k}[/tex]
What is [tex]P(X\ \textless \ 2)[/tex], that is, the probability that she wins fewer than 2 times? Use the binomial probability formula and record your answer to 4 decimal places.
Answer (1)
Calculate the probability of winning 0 times: P ( X = 0 ) = ( 0 5 ) ( 0.2 ) 0 ( 0.8 ) 5 = 0.32768 .
Calculate the probability of winning 1 time: P ( X = 1 ) = ( 1 5 ) ( 0.2 ) 1 ( 0.8 ) 4 = 0.4096 .
Sum the probabilities: P ( X < 2 ) = P ( X = 0 ) + P ( X = 1 ) = 0.32768 + 0.4096 = 0.73728 .
Round to 4 decimal places: P ( X < 2 ) = 0.7373 .
Explanation
Understand the problem and provided data We are given a binomial distribution problem where a person plays a game 5 times, and the probability of winning each game is 0.2. We want to find the probability that she wins fewer than 2 times, which means she wins either 0 times or 1 time. We can use the binomial probability formula to calculate the probabilities of these two scenarios and then add them together. The binomial probability formula is given by: P ( X = k ) = ( k n ) p k ( 1 − p ) n − k where:
n is the number of trials (in this case, the number of games played, which is 5)
k is the number of successful trials (in this case, the number of wins, which can be 0 or 1)
p is the probability of success on a single trial (in this case, the probability of winning a game, which is 0.2)
( k n ) is the binomial coefficient, which represents the number of ways to choose k successes from n trials.
Calculate P(X=0) First, let's calculate the probability of winning 0 times (i.e., losing all 5 games): P ( X = 0 ) = ( 0 5 ) ( 0.2 ) 0 ( 1 − 0.2 ) 5 − 0 Since ( 0 5 ) = 1 and ( 0.2 ) 0 = 1 , we have: P ( X = 0 ) = 1 ⋅ 1 ⋅ ( 0.8 ) 5 P ( X = 0 ) = ( 0.8 ) 5 = 0.32768
Calculate P(X=1) Next, let's calculate the probability of winning exactly 1 time: P ( X = 1 ) = ( 1 5 ) ( 0.2 ) 1 ( 1 − 0.2 ) 5 − 1 Since ( 1 5 ) = 5 , we have: P ( X = 1 ) = 5 ⋅ ( 0.2 ) 1 ⋅ ( 0.8 ) 4 P ( X = 1 ) = 5 ⋅ 0.2 ⋅ ( 0.8 ) 4 = 5 ⋅ 0.2 ⋅ 0.4096 P ( X = 1 ) = 1 ⋅ 0.4096 = 0.4096
Calculate P(X<2) Now, we want to find the probability of winning fewer than 2 times, which is the sum of the probabilities of winning 0 times and winning 1 time: P ( X < 2 ) = P ( X = 0 ) + P ( X = 1 ) P ( X < 2 ) = 0.32768 + 0.4096 = 0.73728 Rounding to 4 decimal places, we get: P ( X < 2 ) = 0.7373
Final Answer The probability that she wins fewer than 2 times is 0.7373.
Examples
Consider a quality control scenario where a factory produces items, and there's a 20% chance an item is defective. If you inspect 5 items, the probability of finding fewer than 2 defective items helps assess the production process's reliability. This calculation is crucial for making informed decisions about process improvements and ensuring product quality.
