1. ## Binomial probability problem

I have a problem working out the answer to this problem using the binomial formula.

Assume an airline operates a short haul flight form Edinburgh to Manchester, there are 16 seats. Experience shows that 15% of people are a 'no show' so they start taking bookings for 20 people.

Using binomial principles we have 20 'trials', P(1 no show) = 0.15, they are all independent.

P(3 no shows) = P(0 no show)+ P(1 no show) + P(2 no shows) + P(3 no shows)
= 0.6477

However I cannot seem to get the answer by using the formula method.
Can anyone help??

Thanks

2. Let be the number of people no show
Then

3. Originally Posted by BGM
Let be the number of people no show
Then

Thanks, that's what I did but I got 0.5101. Maybe it was because I rounded up to much??

4. Originally Posted by ap183

P(3 no shows) = P(0 no show)+ P(1 no show) + P(2 no shows) + P(3 no shows)
= 0.6477
Don't mean to be technical, but the correct way to state it is: P(3 or less no shows). I'm told constantly that its good to speak clearly.

5. Originally Posted by ap183
Using binomial principles we have 20 'trials', P(1 no show) = 0.15, they are all independent.
Maybe what you meant to say was correct but P(1 no show) is not .15 in a sample of 20. The probability of a no show for a single draw is .15 but that doesn't mean that in a sample of 20 there a .15 probability a single person won't show.

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