Question

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

Answer 1

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

Step-by-step explanation:

a) In order to get an even number, you have the 3 different scenarios:

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 doesn't matter

Add them up to get your answer: $\dfrac{1}{16}+\dfrac{1}{16}+\dfrac{1}{16}\quad =\large\boxed{\dfrac{3}{16}}$

b) If one die is a 2 and another is a 3 and the other two dice can be any number, then you have 1 possibility for a 2, 1 possibility for a 3, and 6 possibilities for each of the other two dice.

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