manihar.sidharth wrote:Please help me with this question And please discuss all the probable cases .Please don't give the the solution using 1-P(Not happening the event)
A fair die has sides labeled with 1, 2, 3, 4, 5, and 6 dots. If the die is rolled 4 times, what is the probability that on at least one roll, the die will show a 6?
A)1/6
b)625/1296
c)649/1296
d)671/1296
e)2/3
OA After some discusion
The question is testing your ability to solve the question using the complement: P(at least one 6) = 1 - P(getting no 6's)
tisrar02's solution is great, so I won't repeat it here.
To solve the question without using the complement, we need to see that P(at least one 6) = P(exactly one 6) + P(exactly two 6's) + P(exactly three 6's) + P(exactly four 6's).
Lot's of work here!!!
P(exactly one 6)
Consider one case where we get exactly one 6: six on first roll, non-six on second roll, non-six on third roll, and non-six on forth roll.
To make this solution less cumbersome, we'll denote this outcome as {
6, ~6, ~6, ~6)
So, P(
6, ~6, ~6, ~6) =
(1/6)(5/6)(5/6)(5/6)
We also need to consider the outcome {~6,
6, ~6, ~6)
So, P(~6,
6, ~6, ~6) = (5/6)
(1/6)(5/6)(5/6)
Notice that we can place the 6 in
4 different places to get exactly one 6, so...
P(exactly one 6) = (
4)(1/6)(5/6)(5/6)(5/6) = 500/1296
P(exactly two 6's)
Let's examine one possible outcome: {6, 6, ~6, ~6}
P(6, 6, ~6, ~6) = (1/6)(1/6)(5/6)(5/6)
Of course, {6, 6, ~6, ~6} is just one possible way to get exactly two 6's.
In how many different ways can we get exactly two 6's? Well, in how many different ways can we select 2 places for the 6's?
There are 4 different rolls, and we need to select 2 of them to be 6's. This can be accomplished in 4C2 ways (
6 ways).
So, P(exactly two 6's) = (
6)(1/6)(1/6)(5/6)(5/6) = 150/1296
P(exactly three 6's)
Let's examine one possible outcome: {6, 6, 6, ~6}
P(6, 6, 6, ~6) = (1/6)(1/6)(1/6)(5/6)
In how many different ways can we select 3 places for the 6's?
There are 4 different rolls, and we need to select 3 of them to be 6's. This can be accomplished in 4C3 ways (
4 ways).
So, P(exactly three 6's) = (
4)(1/6)(1/6)(1/6)(5/6) = 20/1296
P(exactly four 6's)
There's only one way to accomplish this.
P(6, 6, 6, 6) = (1/6)(1/6)(1/6)(1/6) = 1/1296
So... P(at least one 6) = (500/1296) + (150/1296) + (20/1296) + (1/1296)
= 671/1296
=
D
Cheers,
Brent
