Assume the random variable $X$ has a binomial distribution with the given probability of obtaining a success. Find the following probability, given the number of trials and the probability of obtaining a success. Round your answer to four decimal places.
$P(X \leq 3), n=6, p=0.3$
Answer (1)
Calculate the probability of X = 0 , 1 , 2 , 3 using the binomial probability formula.
P ( X = k ) = ( k n ) p k ( 1 − p ) n − k , where n = 6 and p = 0.3 .
Sum the probabilities: P ( X ≤ 3 ) = P ( X = 0 ) + P ( X = 1 ) + P ( X = 2 ) + P ( X = 3 ) .
P ( X ≤ 3 ) = 0.1176 + 0.3025 + 0.3241 + 0.1852 = 0.9295 .
Explanation
Understand the problem We are given a binomial distribution with n = 6 trials and a probability of success p = 0.3 . We want to find the probability that X ≤ 3 , which means we need to calculate P ( X = 0 ) + P ( X = 1 ) + P ( X = 2 ) + P ( X = 3 ) .
Recall the binomial probability mass function The probability mass function for a binomial distribution 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 individual probabilities Now, we calculate each term:
For X = 0 :
P ( X = 0 ) = ( 0 6 ) ( 0.3 ) 0 ( 1 − 0.3 ) 6 − 0 = ( 0 6 ) ( 0.3 ) 0 ( 0.7 ) 6 = 1 ⋅ 1 ⋅ ( 0.7 ) 6 = 0.117649
For X = 1 :
P ( X = 1 ) = ( 1 6 ) ( 0.3 ) 1 ( 0.7 ) 5 = 6 ⋅ 0.3 ⋅ ( 0.7 ) 5 = 6 ⋅ 0.3 ⋅ 0.16807 = 0.302526
For X = 2 :
P ( X = 2 ) = ( 2 6 ) ( 0.3 ) 2 ( 0.7 ) 4 = 2 ! 4 ! 6 ! ( 0.3 ) 2 ( 0.7 ) 4 = 2 6 ⋅ 5 ( 0.09 ) ( 0.2401 ) = 15 ⋅ 0.09 ⋅ 0.2401 = 0.324135
For X = 3 :
P ( X = 3 ) = ( 3 6 ) ( 0.3 ) 3 ( 0.7 ) 3 = 3 ! 3 ! 6 ! ( 0.3 ) 3 ( 0.7 ) 3 = 3 ⋅ 2 ⋅ 1 6 ⋅ 5 ⋅ 4 ( 0.027 ) ( 0.343 ) = 20 ⋅ 0.027 ⋅ 0.343 = 0.185220
Sum the probabilities and round the answer Finally, we sum these probabilities to find P ( X ≤ 3 ) :
P ( X ≤ 3 ) = P ( X = 0 ) + P ( X = 1 ) + P ( X = 2 ) + P ( X = 3 ) = 0.117649 + 0.302526 + 0.324135 + 0.185220 = 0.92953 Rounding to four decimal places, we get 0.9295 .
State the final answer Therefore, the probability that X is less than or equal to 3 is approximately 0.9295 .
Examples
Consider a quality control process where items are sampled from a production line. If the probability of an item being defective is 0.3, and we sample 6 items, the probability that 3 or fewer items are defective can be calculated using the binomial distribution. This helps determine if the production process is within acceptable quality limits. For example, if the calculated probability is high, it suggests the process is performing as expected, but if it's low, it might indicate a problem that needs investigation.
