If three fair dice are tossed and the product of the three numbers that appear is even, the probability that the sum of the numbers is also even, is

Question

If three fair dice are tossed and the product of the three numbers that appear is even, the probability that the sum of the numbers is also even, is (A) (1/2) (B) (3/5) (C) (4/7) (D) (5/7)

Tardigrade Admin · Accepted Answer

Product of the three numbers appearing is even

\Rightarrow

atleast one of the throw is even.
Possible cases:

E_{1} \equiv

Exactly one even and 2 odd.

E_{2} \equiv

Exactly two even and 1 odd.

E_{3} \equiv

All three even.
Now

P (E_{1} \cup E_{2} \cup E_{3}) = 1 - P (

all three throws are odd

) = 1 - (\frac{1}{2}) (\frac{1}{2}) (\frac{1}{2}) = \frac{7}{8}

Hence

P (E_{1}

or

E_{3} / E_{1} \cup E_{2} \cup E_{3}) = \frac{P ( E _{1} \cup E _{3} )}{P ( E _{1} \cup E _{2} \cup E _{3} )} = \frac{P ( E _{1} ) + P ( E _{3} ) - P ( E _{1} \cap E _{3} )}{P ( E _{1} \cup E _{3} \cup E _{3} )}

= \frac{3 ( \frac{1}{2} \times \frac{1}{2} \times \frac{1}{2} ) + ( \frac{1}{2} \times \frac{1}{2} \times \frac{1}{2} ) - 0}{\frac{7}{8}} = \frac{4}{8} (\frac{8}{7}) = \frac{4}{7}

Q. If three fair dice are tossed and the product of the three numbers that appear is even, the probability that the sum of the numbers is also even, is

$\frac{1}{2}$

$\frac{3}{5}$

$\frac{4}{7}$

$\frac{5}{7}$

Q. If three fair dice are tossed and the product of the three numbers that appear is even, the probability that the sum of the numbers is also even, is

21​

53​

74​

75​

$\frac{1}{2}$

$\frac{3}{5}$

$\frac{4}{7}$

$\frac{5}{7}$