Question

The probability that a student is accepted to a prestigious college is 0.3. If 5
students from the same school apply. What is the probability of the following.
Use the Binomial Probability formula

a. That at most 2 are accepted?
b. That 2 or more but less than 4?
c. That all 5 are accepted?

Answer 1

For a), you would need to use the binomial formula to find the probability for '1' and '2', and then add them together

Answer 2

Just to be certain, are you familiar with the Binomial Probability formula or used it before?

Answer 3

have not used it... but do I plug in?

Answer 4

http://www.mathwords.com/b/binomial_probability_formula.htm

this is how I would do it correct?

Answer 5

Found this on the web. Might be useful for this question

Answer 6

- p is the probability
- x is the number of successful trials
- n is the total number of events possible

Answer 7

ok awesome thank you so much... let me see if I can work it out

Answer 8

Oh, I didnt see your early reply.
But yes, the formula in that link is correct

Answer 9

Just to correct myself, for a), you would need to find the probability for '0', '1' and '2' .
I forgot to add the zero before. My apologies.

Answer 10

so I would have to use 0.3 and 5 as my numbers for the formula?

Answer 11

Yes, correct. So here is how it would look like:
\[\left(\begin{matrix}n \\ k\end{matrix}\right)*p^k * (1-p)^(n-k)\]

Answer 12

So when k=1,

\[\left(\begin{matrix}5 \\ 1\end{matrix}\right)*0.3^1*(1-0.3)^(5-1)\]

Answer 13

ok working on...

Answer 14

= 4.2 ?

Answer 15

Umm..no. I am not exactly sure where you went wrong. This is what you should get for each one for a):

p(0) = 0.16807
p(1) = 0.36015
p(2) = 0.3087

After adding these, I get a final answer of 0.83692.

Are you sure you used the formula correctly?

Answer 16

Here is how I found 'p(1)' step-by-step using the formula

So we are given that,
k=1
p=0.3
n=5

\[\left( \frac{ 5! }{ 1!(5-1)! } \right)*0.3^1*(1-0.3)^{5-1}\]
\[\left( \frac{ 120 }{ 24 } \right)*0.3*0.7^{4}\]

This gives me an answer of 0.16807 for 'p(1)'.

Hopefully, this can help you figure out where you made the mistake in your calculations. Let me know if you need more explanation.

Answer 17

I see what I did wrong how did you solve it?

Answer 18

Solve which part?

Answer 19

I see how you did it, sorry I am so lost but I am going to write it out again

Answer 20

Thats fine

Answer 21

b(x < 2; 5, 0.3) = b(x = 0; 5, 0.3) + b(x = 1; 5, 0.3) + b(x = 2; 5, 0.3)
b(x < 2; 5, 0.3) = 0.1681 + 0.3601 + 0.3087
b(x < 2; 5, 0.3) = 0.8369

Answer 22

Awesome! Looks correct.

Here is more information to solve other parts of the question:

For b), it says 2 or more but less than 4. That means its asking for probability when
k=2 & 3.

For c), all 5 are accepted meaning find the probability when k=0,1,2,3,4, & 5.

Good Luck!

Answer 23

Thanks I got it!

Answer 24

My pleasure! Glad I could help.