Question

Find the chance of throwing at least one ace in a
simple throw of two dice ?

Answer 1

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& \normalsize \text{ Find the chance of throwing at least one ace in a }\hspace{.33em}\\~\\
& \normalsize \text{ simple throw of two dice ?}\hspace{.33em}\\~\\
\end{align}}\)

Answer 2

ace is related to card problem.... how is ace coming in dice??

Answer 3

ace means one

Answer 4

i m also confused with meaning of ace ,is it 7/36

Answer 5

"The word "ace" comes from the Old French word as (from Latin 'as') meaning 'a unit', from the name of a small Roman coin. It originally meant the side of a die with only one mark, before it was a term for a playing card"

Answer 6

I can provide the solution

Answer 7

(1,1)(1,2)(1,3)(1,4)(1,5)(1,6)

is it 6/36=1/6

Answer 8

no

Answer 9

u said ace =1

the numerator =6 [(1,1)(1,2)(1,3)(1,4)(1,5)(1,6)]

Answer 10

(1,1)(1,2)(1,3)(1,4)(1,5)(1,6)(2,1)(3,1)(4,1)(5,1)(6,1)

Answer 11

See let A denote the statement "no ace in 1st die". and B denote the event "no ace in second die" Then required probability is 1-P(AB) .
Now in the first die, the probability of no ace=5/6 .also in the second die the probability of no ace=5/6. hence required probability =1-P(A)P(B) since the events are independent=1-(5/6)^2=11/36

Answer 12

oh ur right

Answer 13

you could also use

P(1 on die 1 or 1 on die 2)

\[=\frac{1}{6}+\frac{1}{6}-\frac{1}{36}=\frac{11}{36}\]