Question

In four throws with a pair of dices, what is the chance of
throwing a double twice?

Answer 1

\(\large \color{black}{\begin{align}
& \normalsize \text{ In four throws with a pair of dices, what is the chance of}\hspace{.33em}\\~\\
& \normalsize \text{ throwing a double twice? }\hspace{.33em}\\~\\
& a.)\ \dfrac{11}{216} \hspace{.33em}\\~\\
& b.)\ \dfrac{25}{216} \hspace{.33em}\\~\\
& c.)\ \dfrac{35}{216} \hspace{.33em}\\~\\
& d.)\ \dfrac{41}{216} \hspace{.33em}\\~\\
\end{align}}\)

Answer 2

There are 6 ways to get doubles in one throw of a pair of dice.

P(doubles on 1 throw) = 6/36 = 1/6

P(doubles on 1 throw) and P(no doubles on 1 throw) = 1

P(no doubles) = 5/6

Answer 3

Prob ( 2 double throws out of 4) =

C(4,2) * (1/6)^2 * (5/6)^2 =

Answer 4

@mathmath333
doubles could mean double 1, double 2....double 6, so 6/36=1/6.

" In four throws with a pair of dices, what is the chance of throwing a double twice? "

Assuming the question asks "exactly twice".

We know that a throw of 2 dice has 36 outcomes, out of which 6 are doubles.
So the probability of throwing a double in one throw of dice is 1/6.
Since this probability remains constant, and it's a Bernoulli trial (i.e. either success or failure), and the 4 steps of the experiment are random and independent of each other, we can use the binomial distribution, namely
\(P(n,r,p)=nCr~ p^r (1-p)^{n-r}\)
where nCr means n choose r = n!/(r!(n-r)!)
n=number of steps of experiement=4
r=number of successes=2
p=probability of success at each step. (1/6)