A man is known to speak truth 3 out of 4 times. He throws die and reports that it is a 6. The probability that it is actually a 6 is
A) 3/4
B) 5/8
C) 2/5
D) 3/5
E) 4/5
I thought the answer was 3/4*1/6=1/8 using conditional prob and independent events. But it was marked wrong. Can someone pl help...
2006-07-27 07:05:58 · 16 answers · asked by Erin 2 in Science & Mathematics ➔ Mathematics
a) 3/4
he is either liying or telling the truth.
the odds of the die have no place in the formulation of the answer.
2006-07-27 07:07:26 · answer #1 · answered by jimvalentinojr 6 · 0⤊ 1⤋
A)3/4
2006-07-27 08:08:07 · answer #2 · answered by aaaaaaaarrrrrrrr 2 · 0⤊ 0⤋
remember, there are two ways to get a response of "6"
first, he could have rolled a 1 through 5 and lied
second, he could have rolled a 6 and told the truth
you have consider both probabilities
in your original answer, you figured the second part, but neglected the first
we know he reported a six, so we aren't interested in the probability that he would report a 6, we are interested in the probability that given he said 6, what chance is it
the chance he rolled 1-5 and lied is 5/6*1/4=5/24
the chance he rolled 6 and truthed is 1/6*3/4=3/24
the chance that either happened (meaning he said "6") is 8/24
the chance that it is true is the 3/24 of the truth piece, divided by the total possiblities that have him say 6 (8/24) or 3/8
since that isn't the right answer I must have missed something or thought stupid
but if you consider both ways to get a 6 answer and the percentage of those that are truth then you will get there
I got to go to work, lunch hour is over
good luck
2006-07-27 07:28:12 · answer #3 · answered by enginerd 6 · 0⤊ 0⤋
A because the only probability you have to remember is that he tells the truth 3/4 of the time. Its a sort of a trick question, IMO.
Theres no telling how many sides there are to the die, so to assume it is a 6 sided die is logical, but not in the equation, so its an assumption. Never assume anything.
Some dice are 8 sided, some are 10 sided. So the logical probability lies in the mans honesty. We know hes honest 3/4 of the time, but we know nothing about how many sides there are to the die. Actually, die, is the singular, and dice is plural. So you were correct in assuming it was one die, at least. Your logic is great, actually. Keep up the good work. Just never assume anything, lol.
=)
(A) 3/4- his honesty will not become more or less, logically.
2006-07-27 07:12:26 · answer #4 · answered by ♥ Krista ♥ 4 · 0⤊ 0⤋
The probability that it was a 6 AND he reports the truth is:
p1 = 1/6 * 3/4 = 1/8
The probability that it wasn't 6 AND he lied is:
p2 = 5/6 * 1/4 = 5/24
So, he reports 6 with a probability:
p3 = p1 + p2 = 1/8 + 5/24 = 1/3.
Finally the probability that it was actually a 6 is:
P = p1/p3 = (1/8)/(1/3) = 3/8
I think.
2006-07-27 07:43:07 · answer #5 · answered by Dimos F 4 · 0⤊ 0⤋
1 in 6 for the die then 3/4 or 75% probabilty on the 1 in 6
2006-07-27 07:10:29 · answer #6 · answered by Kalahari_Surfer 5 · 0⤊ 0⤋
3/4
2006-07-27 07:07:36 · answer #7 · answered by ConradD 2 · 0⤊ 0⤋
It's a because it doesn't matter what number is on the die he will only answer the truth 3 out of 4 times.
2006-07-27 07:08:30 · answer #8 · answered by Anonymous · 0⤊ 0⤋
It should still be 3/4. The probability of the die throw is irrelavent since the question is only asking about the truth of the statement.
2006-07-27 07:09:44 · answer #9 · answered by rahkokwee 5 · 0⤊ 0⤋
The question is if he is telling the truth, not what the dice say. So it would be 3/4. Your solution attempts to tie the probability of rolling a 6 on the dice, but that is not the question.
2006-07-27 07:08:12 · answer #10 · answered by Blunt Honesty 7 · 0⤊ 0⤋
B. Double the 3 and the 4 and subtract one from the 6 because he has had one chance to lie already.
2006-07-27 07:24:31 · answer #11 · answered by Grist 6 · 0⤊ 0⤋