Question

Three balls are numbered 1, 2, 3 and placed in a bag. A ball is chosen, its number recorded, then the ball placed back in the bag. This process is repeated twice more. The resulting 3 recorded numbers are added.

Bob walks in the room after the three numbers have been recorded and is told only that the sum of the numbers is 6. Bob announces, "Oh, then they must all have been 2's." What is the probability that Bob is correc

Answer 1

The sample space has 27 triplets.
Like 1 1 1, 1 1 2, 1 1 3 etc.
Figure out how many of them add up to 6
call that n
P = 1/n

Answer 2

Or start with the middle digit = 1, 2, or 3
If 1, then only 2 1 3 and 3 1 2 work
If 2, then only 3 2 1 and 1 2 3 work
There's a pattern here!

Answer 3

Forgot 2 2 2

Answer 4

I believe there are 7 triplets with this property.

Answer 5

let me see. but also thanks for the info

Answer 6

Yea there are 7

Answer 7

so p= 1/7

Answer 8

@telliott99 did you see what I wrote?

Answer 9

back

Answer 10

Yes that sounds good.

Answer 11

@telliott99 thanks!

Answer 12

yw

Answer 13

Solution:
The only possible sums that give 6 are
\[1+2+3\]
and
\[2+2+2.\]
Of the \[3^3=27\] total ways to draw three balls in order, there are \[3\cdot 2\cdot 1=6\] ways to draw 1, 2, and 3 in some order. There is also 1 way to draw 2, 2, and 2. There are 7 total ways to draw three balls that sum to 6.

Of these 7 ways, only one of those ways is the draw 2, 2, and then 2. Therefore Bob is only correct in 1 of those 7 cases. The probability that Bob is correct is \[\frac{1}{7}\].