Question

Two fair dice are thrown. Find the probablity
of getting a number divisible by 2 or 4

Answer 1

\(\large \color{black}{\begin{align}
& \normalsize \text{ Two fair dice are thrown. Find the probablity}\hspace{.33em}\\~\\
& \normalsize \text{ of getting a number divisible by 2 or 4}\hspace{.33em}\\~\\
& a.)\ \dfrac{1}{2} \hspace{.33em}\\~\\
& b.)\ \dfrac{3}{4} \hspace{.33em}\\~\\
& c.)\ \dfrac{1}{3} \hspace{.33em}\\~\\
& d.)\ \dfrac{2}{3} \hspace{.33em}\\~\\
\end{align}}\)

Answer 2

The probability of a number divisible by 2
(2,4,6)
=> 3/6 =1/2

Probability of number divisible by 4
(4)
=> 1/6

Final probability
=> (1/2) / (1/6)
=> 2/3

Answer 3

Final probability: W

Answer 4

sorry...typos:-)

Answer 5

Final probability: We have to add
(1/2) + (1/6) = 2/3

Answer 6

but the answer given is a.) 1/2

Answer 7

`two dice ` not just one Jhulian :=))

Answer 8

2 dice, and at least one is divisible by 2 or 4 ? or exactly one?

Answer 9

or is it the sum of both ?

Answer 10

i m also confused about that sum or individual thing that y posted here :)

Answer 11

i see :) lets solve in either way

Answer 12

The correct answer is 1/2 = A

Answer 13

i assumed sum should be divisible by 2 or 4
so
\[(1,1)( 1,3)(1,5) (2,2)(2,4)(2,6)(3,3)(3,5)(4,4)(4,6)((5,5)(6,6)\]
total = 36
12/36 are divisible by 2
\[\color{Red}{( 1,3)(2,2)(2,6)(3,5)(4,4)(6,6)}\]

and 6/36 divisible by 4 \[\frac{ 12 }{ 36}+\color{ReD}{\frac{6}{36}}\]

Answer 14

OMG...My bad :-)