Question

Help with homework please :(///a student makes random guesses on 6 multiple-choice questions. One of the 5 choices in each question is correct. What is the probability that the student will get at most 4 answers right?

Answer 1

okay this question is quite abstract but here goes what i think

A b c d e f
A b c d e f
A b c d e f
A b c d e f
A b c d e f
A b c d e f

imagine all the As are correct
to get all correct, you'd need a probability of (1/5)^6, am i right?

Answer 2

Hi there! I got the part p=1/5 (.2); n=6. What throws me off is the AT MOST 4 is correct.

Answer 3

I am using a graphing calculator: binompdf (6,0.2,x); I need to find x. it is not 4 since the question didn't state EXACTLY 4, but AT MOST 4

Answer 4

yeah i encountered that query too haha

well, i'd suggest finding out P(getting at most 2 wrong), then subtract that probability from 1, otherwise it'd be very tedious

Answer 5

P(X <= 4) = 1 - P(X>4)
= 1 - [P(X=5) + P(X=6)]

Answer 6

6 questions,5choices each
p(0)+p(1)+p(2)+p(3)+p(4) = at most 4 right
or
1-p(5)-p(6)

Answer 7

p(6) is an easy one there is only 1 way to get everything right
1 out of 5^6

Answer 8

p(5)

Answer 9

and just 1 way to pick the 5 other choices

Answer 10

6*(1/5^5)

Answer 11

1-(p(5)+p(6)
1-(6/5^5 + 1/5^6)

Answer 12

here another way to do with probabililty

Answer 13

binomial thm

Answer 14

prob of getting right = 1/5
prob of getting 5 success out of 6 is then

(6Choose5)(1/5)^5*(4/5)

Answer 15

ucan see this also works right away for 6
6Choose6*1/5^6 * 1 = 1/5^6

Answer 16

1-((6Choose5)(1/5)^5*(4/5) + 6Choose6*1/5^6 * 1)

Answer 17

i think i did something wrong while calculaing the prob(5) because its not quite matching up with the other method

Answer 18

\[ P(X\le 4) = 1 - \left[\binom{6}{5} (1/5)^5(4/5)^1 + \binom{6}{6} (1/5)^6(4/5)^0\right]\]