If-two-fair-dice-is-thrown-twice-what-is-the-probability-of-obtaining-an-even-number-

Question Number 217909 by Tawa11 last updated on 23/Mar/25

If two fair dice is thrown twice, what is the
probability of obtaining an even number

Commented by mr W last updated on 24/Mar/25

p = 1 - \frac{2 \times 4}{2^{4}} = \frac{1}{2}

Commented by Unhombre last updated on 23/Mar/25

h e y .

s o w e a r e g i v e n t h a t h a v e t w o d i c e a n d

w e w a n t t h e e v e n n u m b e r s s o :

\frac{3}{6} = \frac{1}{2} = {2, 4, 6}

s i n c e w e h a v e t w o d i c e \Rightarrow \frac{1}{2} \times \frac{1}{2} = \frac{1}{4}

a n d w e h a v e r o l l e d i t t w i c e s o :

\frac{1}{4} \times \frac{1}{4} = \frac{1}{16} o r 6 . 25 %

Commented by Frix last updated on 23/Mar/25

Question not clear . How do we proceed ?

Assuming

n_{1} = d_{1} + d_{2} (1^{st} throw)

n_{2} = d_{1} + d_{2} (2^{nd} throw)

The chance n_{j} is even is p = \frac{1}{2}^{(*)}

\Rightarrow

The chance of both n_{1}, n_{2} are odd is

{(1 - p)}^{2} = \frac{1}{4}

The chance n_{1} or n_{2} are even is

1 - \frac{1}{4} = \frac{3}{4}

(*)

2 1

3 2 \times

4 3

5 4 \times

6 5

7 6 \times

8 5

9 4 \times

10 3

11 2 \times

12 1

Commented by Frix last updated on 23/Mar/25

Dear Unhombre, {you}^{'} re wrong .

The chance to get an even number must

be greater when you throw 4 dices \dots

Commented by Unhombre last updated on 23/Mar/25

H i d e a r F r i x

W e l l i t i s s a y i n g w e h a v e t w o d i c e s a n d w e

t h r o w i t t w i c e s o I d o n t t h i n k I d i d

s o m e t h i n g w r o n g

c o u l d y o u p l e a s e c l a r i f y w h e r e I d i d i t

w r o n g ?

T h a n k y o u

Commented by mr W last updated on 23/Mar/25

i a l s o t h i n k t h a t t h e q u e s t i o n i s n o t

c l e a r . w h a t m e a n s “ o b t a i n i n g a n

e v e n {n u m b e r}^{″} ? t w o d i c e a n d t h r o w n

t w i c e, w e g e t t o t a l l y 4 n u m b e r s . i s

i t v a l i d i f o n l y o n e n u m b e r i s e v e n,

o r a t l e a s t o n e o f t h e s e 4 n u m b e r s

i s e v e n o r a l l 4 n u m b e r s m u s t b e

e v e n ?

Commented by Frix last updated on 23/Mar/25

@ Unhombre

To make it more clear :

We have 2 dices, we want at least one 6

The chance for each dice is \frac{1}{6}

The chance for n o 6 is \frac{5}{6} per dice \Rightarrow the

chance for n o 6 for both dices is

\frac{5}{6} \times \frac{5}{6} = \frac{25}{36}

\Rightarrow the chance for at least one 6 is

1 - \frac{25}{36} = \frac{11}{36} \approx 30 . 56 %

If we have 4 dices, for n o 6 we have

{(\frac{5}{6})}^{4} = \frac{625}{1296}

\Rightarrow the chance for at least one 6 is

1 - \frac{625}{1296} = \frac{671}{1296} \approx 51 . 77 %

Now the chance for an even number is \frac{1}{2}

for 1 dice \Rightarrow the chance for n o even number

cinfusingly is also \frac{1}{2} \Rightarrow n o even number

with 4 dices :

{(\frac{1}{2})}^{4} = \frac{1}{16}

At least one even number

1 - \frac{1}{16} = \frac{15}{16} = 93 . 75 %

Commented by Tawa11 last updated on 23/Mar/25

If it is possible to do everything as a

question . It will be fine .

Find the probability of getting a sum of

even numbers .

Commented by Tawa11 last updated on 23/Mar/25

If two fair dice is thrown twice, what is the
probability of obtaining an even number.
(That is, sum of the numbers is even).
This is what I think.

Commented by mr W last updated on 24/Mar/25

t h r o w n 4 t i m e s, ✓ = s u m i s e v e n

O O O O ✓

O O O E

O O E O

O O E E ✓

O E O O

O E O E ✓

O E E O ✓

O E E E

E O O O

E O O E ✓

E O E O ✓

E O E E

E E O O ✓

E E O E

E E E O

E E E E ✓

Commented by mr W last updated on 24/Mar/25

o n l y o n e n u m b e r i s e v e n :

p = \frac{4}{16} = \frac{1}{4}

a t l e a s t o n e n u m b e r i s e v e n :

p = 1 - \frac{1}{16} = \frac{15}{16}

a l l n u m b e r s a r e e v e n :

p = \frac{1}{16}

Commented by Tawa11 last updated on 24/Mar/25

Thanks for your time sir .

Commented by Tawa11 last updated on 24/Mar/25

Thanks sirs for your time .

Commented by mr W last updated on 25/Mar/25

a r e t h e a n s w e r s r i g h t ?

Commented by Tawa11 last updated on 25/Mar/25

I {don}^{'} t know the answers sir .

But you got the same answer

as sir frix .

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