Tossing a die will have 6 possible outcomes. Those are having sides that are number 1 to 6. The sample space of tossing 3 dice is equal to 6³ which is equal to 216. Now for the calculation of probabilities,
P(two 5s) = (1 x 1 x 5)/216
As we have to have the 5 in the die for two times, then for the 1 time, we can have all other numbers except 5. The answer is 5/216.
P(three 5s) = (1 x 1 x 1)/216 = 1/216
P(one 5 or two 5s) = (1 x 5 x 5)/216 + (1 x 1 x 5)/216 = 5/36
Hoi!
To solve this, first plug in the values for x and y.
x = 2, so anywhere you see x, put 2 in its place.
y = 3, so anywhere you see y, put 3 in its place.
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4 × 2 = 8

× 3 =
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
=
is your answer.
I’m pretty sure a is always and b is sometimes.
After reading the problem, you are often asked to solve it.
They were born 2336 years apart.
384+1952=2336