If [tex]$X$[/tex] is a binomial random variable in an experiment with 5 trials, and in which [tex]$P($[/tex] success [tex]$)=0.3, P(X \ \textgreater \ 3) = $[/tex]
0.996
0.031
0.004
Answer (1)
Calculate P ( X = 4 ) using the binomial probability formula: P ( X = 4 ) = ( 4 5 ) ( 0.3 ) 4 ( 0.7 ) 1 = 0.02835 .
Calculate P ( X = 5 ) using the binomial probability formula: P ( X = 5 ) = ( 5 5 ) ( 0.3 ) 5 ( 0.7 ) 0 = 0.00243 .
Add the probabilities: 3) = P(X=4) + P(X=5) = 0.02835 + 0.00243 = 0.03078"> P ( X > 3 ) = P ( X = 4 ) + P ( X = 5 ) = 0.02835 + 0.00243 = 0.03078 .
Round the result to three decimal places: 0.031 .
Explanation
Understand the problem We are given a binomial random variable X with n = 5 trials and probability of success p = 0.3 . We want to find 3)"> P ( X > 3 ) , which means the probability that X is greater than 3.
Express P(X > 3) in terms of P(X=4) and P(X=5) Since X can only take integer values, 3)"> P ( X > 3 ) is the same as P ( X = 4 or X = 5 ) . Because X = 4 and X = 5 are mutually exclusive events, we have 3) = P(X=4) + P(X=5)"> P ( X > 3 ) = P ( X = 4 ) + P ( X = 5 )
State the binomial probability mass function The probability mass function for a binomial random variable is given by P ( X = k ) = ( k n ) p k ( 1 − p ) n − k where ( k n ) = k ! ( n − k )! n ! is the binomial coefficient.
Calculate P(X=4) We calculate P ( X = 4 ) using the formula: P ( X = 4 ) = ( 4 5 ) ( 0.3 ) 4 ( 0.7 ) 5 − 4 = ( 4 5 ) ( 0.3 ) 4 ( 0.7 ) 1 ( 4 5 ) = 4 ! 1 ! 5 ! = ( 4 × 3 × 2 × 1 ) ( 1 ) 5 × 4 × 3 × 2 × 1 = 5 P ( X = 4 ) = 5 × ( 0.3 ) 4 × ( 0.7 ) = 5 × 0.0081 × 0.7 = 5 × 0.00567 = 0.02835
Calculate P(X=5) We calculate P ( X = 5 ) using the formula: P ( X = 5 ) = ( 5 5 ) ( 0.3 ) 5 ( 0.7 ) 5 − 5 = ( 5 5 ) ( 0.3 ) 5 ( 0.7 ) 0 ( 5 5 ) = 5 ! 0 ! 5 ! = 1 P ( X = 5 ) = 1 × ( 0.3 ) 5 × 1 = ( 0.3 ) 5 = 0.00243
Calculate P(X > 3) and round the answer Finally, we add the probabilities: 3) = P(X=4) + P(X=5) = 0.02835 + 0.00243 = 0.03078"> P ( X > 3 ) = P ( X = 4 ) + P ( X = 5 ) = 0.02835 + 0.00243 = 0.03078 Rounding to three decimal places, we get 0.031 .
Examples
Consider a quality control process where you inspect 5 items from a production line, and each item has a 30% chance of being defective. Calculating 3)"> P ( X > 3 ) tells you the probability of finding more than 3 defective items in your sample. This helps you assess the overall quality of the production line and make informed decisions about whether to adjust the manufacturing process.
