Screenshot is from dedicated data session 4. I’m scoring 160+ on Quant consistently, but I’m horrible at setting up probability/combo problems.
When I try to solve this, I set it up like this vvvvvv I’m writing the following realizing that parts are incorrect.
(3 instances) –– –– ––
(3 cases that work) 5 / 5 / 1, 3, 5
5 / 1, 3, 5 / 5
1, 3, 5 / 5 / 5
Each line has odds of 1/6 * 1/6 * 1/2 = 1/72
Therefore the three cases equal 3/72 or 1/24
Where have I gone wrong?!
So you need the product to be an (odd int + div by 25)
To be divisible by 25 , we need atleast 2 5’s : thus , whatever our dice rolls will be, they should contain atleast 2 five:
5 x 5 x z
Now, z can be any number from 1 to 6 but we want our result to be odd, thus , only 1 ,3, 5 will work
Hence, we should roll :
5x5x1
5x5x3
5x5x5
But look at (5x5x1) and (5x5x3), we can also write them as (1x5x5),(3x5x5) and(5x1x5),(5x3x5):-> Because its a combination problem, so, in total 3 ways to write each of those two cases
Probability = desired cases/ total cases
desired cases = (5x5x1),(5x5x3),(5x5x5),(1x5x5),(3x5x5),(5x1x5),(5x3x5) = 7
Total cases = Max outcomes in a dice[1,2,3,4,5,6=6] .i.e = 6 , and since we have 3 dice = 6^3
Ans = 7/(6^3)
P.S. to get better at probability questions , you always need to figure out desired cases and total cases . Now, when u practice try to think in terms of desired cases and total cases.
@HoldMyBeer
Instead of the approach mentioned by you, can’t we directly use the logic that
for Dice 1
Prob for getting a 5 is 1/6 Similarly for Dice 2 it is 1/6 and for Dice 3 to have 1,3,5 it is 1/2
getting us with 1/72 but also making sure that the same can be written as (1x5x5…) giving us 7 cases with 3 dices
This still makes the Probability as 7/72
Could you help me understand the flaw in this method? Your method seems absolutely right but I still can’t understand where am I getting wrong!
Probability for rolling a (5,5,5) : \frac{1}{6} \times\frac{1}{6} \times\frac{1}{6} =\frac{1}{216}
Probability for rolling a (5,5,1) : \frac{1}{6} \times\frac{1}{6} \times\frac{1}{6} but as the order matter we can have one in \frac{3!}{(3-1)!}=3 ways, thus, \frac{1}{6} \times\frac{1}{6} \times\frac{1}{6} \times3 = \frac{3}{216}
same for (5,5,3) and then you add all the probabilities
when you do \frac{1}{2} \times 3 because of 3 dice , you’re counting {5,5,5,} occurring 3 times as well which is not the case, you can either set it like {5,5,5} :-> condition1 and {1 or 3 } as condition 2 and then add all the probabilities
So ,
\frac{1}{6} \times \frac{1}{6} \times\frac{1}{6} = \frac{1}{216}
for either 1 or 3
\frac{1}{6} \times \frac{1}{6} \times \frac{2}{6} \times 3 = \frac{1}{36}
Adding:
\frac{1}{216} + \frac{1}{36} = \frac{7}{216}
Aaah! Got it now! Thanks Much.
