Binomial probability problem

**ap183** · 08-04-2010 08:12 AM

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

**BGM** · 08-04-2010 09:12 AM

Let $N$ be the number of people no show
Then $N \sim \rm {Binomial} (20, 0.15)$

$\Pr\{N \leq 3\} = \Pr\{N = 0\} + \Pr\{N = 1\} + \Pr\{N = 2\} + \Pr\{N = 3\}$

$= \binom {20} {0} (0.15^0)(0.85^{20}) + \binom {20} {1} (0.15^1)(0.85^{19})$ $+ \binom {20} {2} (0.15^2)(0.85^{18}) + \binom {20} {3} (0.15^3)(0.85^{17})$

$\approx 0.0387595310845143 + 0.1367983450041683 + 0.2293384019187528$ $+ 0.2428288961492676$

$\approx 0.647725174156703$

**ap183** · 08-04-2010 10:34 AM

Originally Posted by BGM

Let $N$ be the number of people no show
Then $N \sim \rm {Binomial} (20, 0.15)$

$\Pr\{N \leq 3\} = \Pr\{N = 0\} + \Pr\{N = 1\} + \Pr\{N = 2\} + \Pr\{N = 3\}$

$= \binom {20} {0} (0.15^0)(0.85^{20}) + \binom {20} {1} (0.15^1)(0.85^{19})$ $+ \binom {20} {2} (0.15^2)(0.85^{18}) + \binom {20} {3} (0.15^3)(0.85^{17})$

$\approx 0.0387595310845143 + 0.1367983450041683 + 0.2293384019187528$ $+ 0.2428288961492676$

$\approx 0.647725174156703$

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

**Link** · 08-04-2010 10:55 AM

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.

**Dason** · 08-04-2010 10:57 AM

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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