gayathri wrote:
What is the probability of getting a sum of 8 or 14 when rolling 3 dice simultaneously?
A. 1/6
B. 1/4
C. 1/2
D. 21/216
E. 32/216
hey
pushpitkc , can you please let me know if my reasoning and approach is correct ?
if not please identify the flaws:)
thank you and have a great weekend
Number of Ways to get 8 (not to get confused and or not to miss some number, start listing number in increasing order
)
116
125
134
143
152
161
now i see that there are 6 ways, but since each combination has 3 numbers, i can reranrange each combination in 3 ways
Hence 6 * 3 = 18
to calculate number of ways to get sum 14, i didnt have to enumerate favorite outcomes, because logically it must be the same number of ways as is the case with 8
Hence 6 * 3 = 18
So total number of favoroble outcomes is 18+18 = 36
Total Number of Ways: so we have three dices
choosing any number from first dice: is 1/6
choosing any number from second dice: is 1/6
choosing any number from third dice is 1/6
hence total number of way \(\frac{1}{6} * \frac{1}{6} * \frac{1}{6}\) = \(\frac{1}{216}\)
probability of getting a sum of 8 or 14 when rolling 3 dice simultaneously is \(\frac{36}{216}\) i.e =\(\frac{1}{6}\)
hey
pushpitkc are you there ?
need your help clearing confusions
I have a doubt regarding total number of ways I wrote above, in one of the posts I saw that total number of ways is 6*6*6
ok dice has 6 dfferent numbers, and we have three dices: when we say we can get any number in 6 ways, why we write 6*6*6 and not 6! *6!*6! ? can you explain the difference? and why multiplying \(\frac{1}{6}\) by itself three times is incorrect ?Hello
niks18 may be you can help me to clear my confusion. see the highlighted part above
thank you