Find the probability distribution of the number of successes in two tosses of a die,where a success is defined as 'six appears on at least one die'.

(N/A) Let $Y$ be the random variable representing the number of successes in two tosses of a die. $A$ success is defined as 'six appears on at least one die'.
The total number of outcomes when two dice are tossed is $6 \times 6 = 36$.
Let $S$ be the event that a six appears on a die,and $F$ be the event that a six does not appear. $P(S) = \frac{1}{6}$ and $P(F) = \frac{5}{6}$.
$P(Y=0)$ is the probability that a six does not appear on either die. This corresponds to the outcomes where neither die shows a six: $P(Y=0) = \frac{5}{6} \times \frac{5}{6} = \frac{25}{36}$.
$P(Y=1)$ is the probability that a six appears on at least one die. This corresponds to the outcomes where at least one die shows a six: $P(Y=1) = 1 - P(Y=0) = 1 - \frac{25}{36} = \frac{11}{36}$.
Wait,let us re-evaluate the definition of success. If success is 'six appears on at least one die',then in two tosses,we can have either $0$ successes or $1$ success (since the event 'six appears on at least one die' is a single outcome event per trial pair). However,if we treat the two tosses as independent trials,the probability distribution is:

$Y$	$0$	$1$
$P(Y)$	$\frac{25}{36}$	$\frac{11}{36}$