Probability question

auonline · #1 $Old$ June 6th, 2009, 05:29 PM

A new drug is being tested on 12 patients with a type of cancer. It is thought to cure patients with 0.7 probability. What is the probability that:

d) fewer than 4 patients will be cured?
e) more than 7 patients will be cured?

How can I do this both manually and in the calculator?
Thanks

e^(i*pi) · #2 $Old$ June 6th, 2009, 05:35 PM

Quote:

Originally Posted by auonline $View Post$

A new drug is being tested on 12 patients with a type of cancer. It is thought to cure patients with 0.7 probability. What is the probability that:

d) fewer than 4 patients will be cured?
e) more than 7 patients will be cured?

How can I do this both manually and in the calculator?
Thanks

Well there is a 0.3 chance they won't be cured. (Assuming this is mutually exclusive and they are either cured or not cured)

The chance that n patients will be cured is given by

$0.7^n \times 0.3^{12-n}$

In part d) n = 3 because fewer than 4 means 0,1,2 or 3

In part e) n = 7

The 12-n comes from the total minus the number that are cured

auonline · #3 $Old$ June 6th, 2009, 05:42 PM

For part d), do I add the probabilities of n = 0 , 1 , 2 and 3 all together or is there a quicker way via graphics calculator?

e) n = 7? what about n = 8, n = 9 etc.???

Thanks :s

These are meant to be binomial probability questions if that helps

e^(i*pi) · #4 $Old$ June 6th, 2009, 05:57 PM

Quote:

Originally Posted by auonline $View Post$

For part d), do I add the probabilities of n = 0 , 1 , 2 and 3 all together or is there a quicker way via graphics calculator?

e) n = 7? what about n = 8, n = 9 etc.???

Thanks :s

These are meant to be binomial probability questions if that helps

The binomial distribution is beyond my knowledge I'm afraid, I was using powers, the idea being that they were independent events and so each one would be either cured with 0.7 or not cured with 0.3
You could add them all up, not sure how you'd use a calculator to plot a graph though.

mr fantastic · #5 $Old$ June 6th, 2009, 07:12 PM

Quote:

Originally Posted by auonline $View Post$

A new drug is being tested on 12 patients with a type of cancer. It is thought to cure patients with 0.7 probability. What is the probability that:

d) fewer than 4 patients will be cured?
e) more than 7 patients will be cured?

How can I do this both manually and in the calculator?
Thanks

Let X be the random variable number of patients cured.

X ~ Binomial(n = 12, p - 0.7)

d) Calculate $\Pr(X < 4) = \Pr(X \leq 3)$ .

e) Calculate $\Pr(X > 7) = \Pr(X \geq 8)$ .

Doing this manually requires using the formula $\Pr(X = x) = {12 \choose x} (0.7)^x (0.3)^{12 - x}$ . eg. for d) you have to calculate Pr(X = 0) + Pr(X = 1) + Pr(X = 2) + Pr(X = 3) using this formula.

How you do it using a calculator will greatly depend on the type of calculator you own. I suggest you read the User Manual (or find and read the user manual on-line if you've done what every student I've ever met has done - lost your copy that came with the calculator).

The arithmetic itself can easily be done using a scientific calculator but I doubt that's what you meant ....