# Math Help - Binomial Distribution, balls and questions

1. ## Binomial Distribution, balls and questions

I have 2, two part questions.

1. A box contains 3 yellow and 5 black balls. A ball is drawn and then replaced. This process is continued for a total of 5 draws. what is the probability of drawing:
a. exactly 4 black balls
b. at most 4 black balls

2. A test contains 20 multiple choice questions each with 3 choices. If a student guesses every answer, what is the probability he will get:

Thanks
~G

2. 1. A box contains 3 yellow and 5 black balls. A ball is drawn and then replaced. This process is continued for a total of 5 draws. what is the probability of drawing:
a. exactly 4 black balls
$C(5,4)(\frac{5}{8})^{4}(\frac{3}{8})^{1}$

b. at most 4 black balls
EDIT: thanks to Janes cognizance. I fix. See later post.

2. A test contains 20 multiple choice questions each with 3 choices. If a student guesses every answer, what is the probability he will get:
$(\frac{2}{3})^{20}$

$C(20,5)(\frac{1}{3})^{5}(\frac{2}{3})^{15}$

Thanks
~G[/quote]

3. Originally Posted by galactus
Subtract the above solution from 1.
It’s actually 1 − P(exactly 5 black) − P(exactly 6 black) − P(exactly 7 black) − P(exactly 8 black).

4. But there are only 5 blacks altogether. Nonetheless, I had a brain fart. I was thinking of the 'at least one' case.

$\sum_{k=0}^{4}C(5,k)(\frac{5}{8})^{k}(\frac{3}{8}) ^{5-k}=\frac{29643}{32768}\approx{.9046}$

or

$1-(\frac{5}{8})^{5}= \frac{29643}{32768}\approx{.9046}$

5. ## thanks

I really appreciate both of you taking the time to help me.