Please see the two examples below and their solutions posted on this forum-majority of the responses were for the same answers. i.e for the 1st question majority explained their answer as D-1/2 and for the second question the answer as D-21/45.
I guess the solution for the 1st answer is wrong here-the numerator should be 1*4c1 and not 1+4c1.this is where the problem lies..guys please tell me the right way to approach these problems.
16) A basket has 5 apples, one is spoiled. If Henry picked 2 apples simultaneously and at random, what is probability that the 2 selected apples will include the spoiled one.
a. 1�5
b. 2�10
c. 2�5
d. 1�2
e. 3�5
Solution-
Here as per my understanding this question can be solved in this way:
Henry picks 2 apples from 5 apples in the basket so total possible ways is:5c2= 10.
Possibility that 2 selected apples include spoiled one is (Probality of getting one spolied one + Probability of getting one good one from the 4 apples)
this probability will be 1 + 4c1 = 5
Hence the probability of getting one spoiled among the two apples picked randomly is 5/10 = 1/2.
Hi, can someone help me with this problem- it seems simple, but I must be mistaken in my logic. Its from the Powerprep software Exam 2.
A certain basket contains 10 apples, 7 of which are red and 3 of which are green. If 3 different apples are to be selected at random from the basketbal, what is the probability that 2 of the apples selected will be red and 1 will be green?
A) 7/40
B) 7/20
C) 49/100
D) 21/40
E) 7/10
I thought it would just be 7/10 * 6/9 * 3/8, but thats wrong.
I'll post the answer after some discussion -- thanks.
Solution----
I think it's pretty simple...Let's C if I am right!
7C2 * 3C1 / 10C3
= 21/40
= Choice D
Cheers
