Solving Binomial Dist. Exercise with n=64 & p=0.2 - Help Needed!

[wiz0r]

The exercisie is in the attachment.

I am trying to solve it using a Binomial distribution, using n = 64, and p = 0.2.

P(X=1) =

(64) (0.2)^1 (0.8)^63
( 1)

and, I get that P(X=1) = 1.00434 * 10^-5

However, this does not seems correct, can anyone help me here?? I need to submit this in 4hours, so please help me. Thanks.

 

[wiz0r]

Could the answer be: 0.030260441?

 

[Hurkyl]

How did you calculate that number?

 

[Mark44]

Seems OK to me. As you described it, you're looking for the probability of 1 success in 64 trials, where each trial has a 0.2 probability of success.

Another way to think about this is that you have a jar with 5 balls in it, with one white ball and four black balls. For each trial, you reach in (without looking) and pick out a ball and record its color, then put the ball back in. If you do this 64 times, what's the probability that you get 1 white ball out of 64 trials? It's pretty low.

 

[Mark44]

The exercisie is in the attachment.

I am trying to solve it using a Binomial distribution, using n = 64, and p = 0.2.

P(X=1) =

(64) (0.2)^1 (0.8)^63
( 1)

and, I get that P(X=1) = 1.00434 * 10^-5

However, this does not seems correct, can anyone help me here?? I need to submit this in 4hours, so please help me. Thanks.

Now that I can see your attachment, what you calculated seems correct for P(X = 1), but this isn't the probability that the problem asks for, which is the probability that at least one passenger will get bumped.

 

[wiz0r]

Thanks for the input so far guys:

As for my last attempt:

After I careful read the problem, it says "at least one passenger..", so I assumed that it was either 1 passager needs to be "bumped", or 2, or 3, or 4, or 5, or 6.

So, I added the probabilities of:

P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) + P(X = 5) + P(X = 6)

And that's how I got the 0.030260441.

 

[wiz0r]

Yes, Mark44! Is it correct now??

 

[Mark44]

Looks OK, assuming you calculated the individual probabilities correctly. For example, for P(X = 2) you have ₆₄C₂*(.2)²*(.8)⁶², right? (That number at the beginning works out to 64*63/2.)

 

[wiz0r]

Thank you, Mark. I did it with a formula using Excel, so I'm pretty sure it's correct. Thanks a lot!

 

[wiz0r]

Actually, I had an error there, the correct answer is: 0.027605266.

My bad.

 

[Hurkyl]

Anyways, now that you've worked it out, it might interest you to know there's a cute shortcut -- what is the probability that nobody gets bumped?