91Ó°ÊÓ

From the data see that the two dice are tossed then the example space of the events is as per the following

consider $S$is the event that addresses the example space.

$S=\left\{\begin{array}{l}(1,1),(1,2),(1,3),(1,4),(1,5),(1,6),\\ (2,1),(2,2),(2,3),(2,4),(2,5),(2,6),\\ (3,1),(3,2),(3,3),(3,4),(3,5),(3,6),\\ (4,1),(4,2),(4,3),(4,4),(4,5),(4,6),\\ (5,1),(5,2),(5,3),(5,4),(5,5),(5,6),\\ (6,1),(6,2),(6,3),(6,4),(6,5),(6,6)\end{array}\right\}$

The sum of numbers of two dice are,

Consider $E$is the event that addresses something like one of the pair of fair dice lands on $6$

Consider $F$is the event that addresses the amount of the number on the two dice.

The amount of the dice might be any of the numbers beginning from $2$to $12$

In any case, to get somewhere around one of the two numbers like $6$the total beginnings from$7$, not from $2$.

That is the probability of getting something like one of the two numbers as $6$given that the aggregates are$2,3,4,5,$and $6$ are zeros.

Subsequently,

$P(Eâˆ\pounds F=2)=0$
$P(Eâˆ\pounds F=3)=0$

$P(Eâˆ\pounds F=4)=0$

$P(Eâˆ\pounds F=5)=0$

$P(Eâˆ\pounds F=6)=0$

Work out the probability that something like one of the pair of fair dice lands on $6$given that the sum on the two dice is $7$.

The ideal number of something like one of the pair of fair dice lands on six and the sum of two numbers on the dice is $7$as per the following:

$Eâˆ\copyright F=\left\{\right(1,6),(6,1\left)\right\}$

In this manner, the probability of something like one of the pair of fair dice lands on six, and the sum of two numbers on the dice is $7$ is,

$P(Eâˆ\copyright F=7)=\frac{2}{36}$

The complete number of cases that the sum of the two numbers on the dice is $7$ is as per the following:

$F=\left\{\right(1,6),(2,5),(3,4),(4,3),(5,2),(6,1\left)\right\}$

In this way, the probability that the sum of the two numbers on the dice is$7$is,

$P(F=7)=\frac{6}{36}$

Hence, the conditional probability is,

$P(Eâˆ\pounds F=7)=\frac{P(Eâˆ\copyright F=7)}{P(F=7)}$

localid="1647597381785" $=\frac{{\displaystyle \frac{2}{36}}}{{\displaystyle \frac{6}{36}}}$

localid="1647597414744" $=\frac{2}{6}$

Work out the probability that something like one of the pair of fair dice lands on $6$given that the sum on the two dice is $8$.

The ideal number of something like one of the pair of fair dice lands on six and the sum of two numbers on the dice is $8$is as per the following:

$Eâˆ\copyright F=\left\{\right(2,6),(6,2\left)\right\}$

In this manner, the probability of something like one of the pair of fair dice lands on six, and the sum of two numbers on the dice is $8$is,

$P(Eâˆ\copyright F=8)=\frac{2}{36}$

The complete number of cases that the sum of the two numbers on the dice is $8$ is as per the following:

$F=\left\{\right(2,6),(3,5),(4,4),(5,3),(6,2\left)\right\}$

In this way, the probability that the sum of the two numbers on the dice $8$ is,

$P(F=8)=\frac{5}{36}$

Hence, the conditional probability is,

$P(Eâˆ\pounds F=8)=\frac{P(Eâˆ\copyright F=8)}{P(F=8)}$

$=\frac{{\displaystyle \frac{2}{36}}}{{\displaystyle \frac{5}{36}}}$

localid="1647602472318" $=\frac{2}{5}$

Work out the probability that something like one of the pair of fair dice lands on given that the sum on the two dice is.

The ideal number of something like one of the pair of fair dice lands on six and the sum of two numbers on the dice is $9$as per the following:

$Eâˆ\copyright F=\left\{\right(3,6),(6,3\left)\right\}$

In this manner, the probability of something like one of the pair of fair dice lands on six, and the sum of two numbers on the dice is$9$,

$P(Eâˆ\copyright F=9)=\frac{2}{36}$

The complete number of cases that the sum of the two numbers on the dice is $9$is as per the following:

$F=\left\{\right(3,6),(4,5),(5,4),(6,3\left)\right\}$

In this way, the probability that the sum of the two numbers on the dice is $9$is,

$P(F=9)=\frac{4}{36}$

Hence, the conditional probability is,

$P(Eâˆ\pounds F=9)=\frac{P(Eâˆ\copyright F=9)}{P(F=9)}$

$=\frac{\frac{2}{36}}{\frac{4}{36}}$

$=\frac{2}{4}$

Work out the probability that something like one of the pair of fair dice lands on given that the sum on the two dice is.

The ideal number of something like one of the pair of fair dice lands on six and the sum of two numbers on the dice is $10$s as per the following:

$Eâˆ\copyright F=\left\{\right(4,6),(6,4\left)\right\}$

In this manner, the probability of something like one of the pair of fair dice lands on six, and the sum of two numbers on the dice is$10$,

$P(Eâˆ\copyright F=10)=\frac{2}{36}$

The complete number of cases that the sum of the two numbers on the dice is $10$is as per the following:

$F=\left\{(4,6),(5,),(6,4)\right\}$

In this way, the probability that the sum of the two numbers on the dice is $10$is,

Hence, the conditional probability is,

$P(Eâˆ\pounds F=10)=\frac{P(Eâˆ\copyright F=10)}{P(F=10)}$

$=\frac{{\displaystyle \frac{2}{36}}}{{\displaystyle \frac{3}{36}}}$

$=\frac{2}{3}$

Work out the probability that something like one of the pair of fair dice lands on $6$given that the sum on the two dice is $11$.

The ideal number of something like one of the pair of fair dice lands on six and the sum of two numbers on the dice is $11$is as per the following:

$Eâˆ\copyright F=\left\{(5,6),(6,5)\right\}$

In this manner, the probability of something like one of the pair of fair dice lands on six, and the sum of two numbers on the dice is $11$is,

$P(Eâˆ\copyright F=11)=\frac{2}{36}$

The complete number of cases that the sum of the two numbers on the dice is$11$is as per the following:

$F=\left\{(5,6),(6,5)\right\}$

In this way, the probability that the sum of the two numbers on the dice is $11$ is,

$P(F=11)=\frac{2}{36}$

Hence, the conditional probability is,

$P(E/F=11)=\frac{P(Eâˆ\copyright F=11)}{P(F=11)}$

$=\frac{{\displaystyle \frac{2}{36}}}{{\displaystyle \frac{2}{36}}}$

$=1$

Work out the probability that something like one of the pair of fair dice lands on $6$given that the sum on the two dice is$12$.

The ideal number of something like one of the pair of fair dice lands on six and the sum of two numbers on the dice is $12$ is as per the following:

$Eâˆ\copyright F=\left\{\right(6,6\left)\right\}$

In this manner, the probability of something like one of the pair of fair dice lands on six, and the sum of two numbers on the dice is $12$ is,

$P(Eâˆ\copyright F=12)=\frac{1}{36}$

The complete number of cases that the sum of the two numbers on the dice is $12$ is as per the following:

$F=\left\{\right(6,6\left)\right\}$

In this way, the probability that the sum of the two numbers on the dice is $12$is,

$P(F=11)=\frac{1}{36}$

Hence, the conditional probability is,

$P(Eâˆ\pounds F=12)=\frac{P(Eâˆ\copyright F=12)}{P(F=12)}$

$=\frac{\frac{1}{36}}{\frac{1}{36}}$

$=1$

The conditional probability of $\left(E\right|F=2),(E|F=3),\left(E\right|F=4),(E|F=5)and\left(E\right|F=6)$are Zero.

The conditional probability of $P\left(E\right|F=7)$is $\frac{2}{6}$

The conditional probability of $P\left(E\right|F=8)$is$\frac{2}{5}$

The conditional probability of $P\left(E\right|F=9)$is $\frac{2}{4}$

The conditional probability of $P\left(E\right|F=10)$is $\frac{2}{3}$$\frac{2}{3}$

The conditional probability of $P\left(E\right|F=11)$is $1$width="10">1

The conditional probability of $P\left(E\right|F=12)$is $1$