Zaid has a peculiar pair of four-sided dice. When he rolls the dice, the probability of any particular outcome is proportional t

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2 months ago

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Zaid has a peculiar pair of four-sided dice. When he rolls the dice, the probability of any particular outcome is proportional t

o the sum of the results of each die, i-e, for any outcome (i, j), 1 i, j 4, P (i, j) = (i + j) p, where p is the constant of proportionality. All outcomes that result in a particular sum are equally likely. (a) What is the probability of the sum being even? (b) What is the probability of Bob rolling a 2 and a 3, in any order?

Mathematics

1 answer:

AnnZ [12.3K] · Answer 1 · 2026-03-08T13:03:49+03:00

Answer: a) $\bold{\dfrac{3}{16}}$ b) $\bold{\dfrac{1}{36}}$

Step-by-step explanation:

a) To achieve an even total, there are 3 possible combinations:

1) Even, Even, Even, Even $\dfrac{3\times 3\times 3\times 3}{6^4} \quad = \dfrac{3^4}{6^4}\quad =\dfrac{1}{16}$

2) Even, Even, Odd, Odd $\dfrac{3\times 3\times 3\times 3}{6^4} \quad = \dfrac{3^4}{6^4}\quad =\dfrac{1}{16}$

3) Odd, Odd, Odd, Odd $\dfrac{3\times 3\times 3\times 3}{6^4} \quad = \dfrac{3^4}{6^4}\quad =\dfrac{1}{16}$

Order is irrelevant

Summing these yields your final result: $\dfrac{1}{16}+\dfrac{1}{16}+\dfrac{1}{16}\quad =\large\boxed{\dfrac{3}{16}}$

b) If one die shows a 2 and another a 3, while the remaining two can show any digits, there’s only one way to get a 2, one way for a 3, and six potential numbers for each of the other two dice.

$\dfrac{1\times 1\times 6\times 6}{6^4}\quad =\dfrac{1}{6^2}\quad =\large\boxed{\dfrac{1}{36}}$