# Thread: Multiple Choice Exam & Jelly Beans in a Bag

1. ## Multiple Choice Exam & Jelly Beans in a Bag

Probability drives me crazy
I couldn't figure out how it solves

Multiple Choice Exam
Each question on a 5 questions multiple choice examination has 4 choices,only one of which is correct . If a student answers each question on a random fashion, what's the probability that the student answers exactly 2 question incorrectly ?
# $(S)=1024 (4^5)$
# $(E)=10 (5C3)$ I said if the student answers 2 question incorrectly is the same as if he answers ignore any 2 question
$P(E)= 10/1024$

tell me how to solve it plz

BUT the right answer is ( $45/512$)

Jelly Beans in a Bag

Abag contains 4 red and 6 green jelly beans.
a) if 2 jelly beans are randomly selected in succession with replacement, determine the probability that both are red.
b)if the selection is made without replacement .determine the probability that both are red.

#( $S)=24$
a)# $(E)= 4c2
P(E)= 6/24=1/4
$

b)#( $E)= 4p2
P(E)= 12/24=1/2
$

$a)4/25
$

$
b)2/15
$

Tell me please how to solve them

2. The first one is solved by realizing that the student gets 3 correct and two incorrect, that the probability of a correct answer on any one question is 1/4, and the probablilty of an incorrect answer is 3/4. The right and wrong answers can be arranged in any one of 5C2 ways. So you get:

5C2*(1/4)^3 * (3/4)^2 = 10*9/1024 = 45/512.

For the second problem: there are 4 reds and 10 total, so the probability of pulling a red on the first draw is 4/10. If you put that red one back and draw again, the probablity of drawing red a second time is also 4/10. But if you don't replace it, then for the second draw there are only 3 reds out of the 9 that are left, so the probability of a second red is 3/9. Putting it together:
a) prob of two reds in a row with replacement = 4/10 * 4/10 = 4/25.
b) prob of two reds in a row without replacement = 4/10 * 3/9 = 2/15.