Q.3.4 - A First Course in Probability

Question

What is the probability that at least one of a pair of fair dice lands on $6$, given that the sum of the dice is $i, i = 2, 3, ... , 12$?

Step-by-Step Solution

Answer

Conditional Probabilities of

$P\left(E\right|F=2)=0$

$P\left(E\right|F=3)=0$

$P\left(E\right|F=4)=0$
$P\left(E\right|F=5)=0$

$P\left(E\right|F=6)=0$

$P\left(E\right|F=7)=\frac{2}{6}$

$P\left(E\right|F=8)=\frac{2}{5}$

$P\left(E\right|F=9)=\frac{2}{4}$

$P\left(E\right|F=10)=\frac{2}{3}$

$P\left(E\right|F=11)=1$

$P\left(E\right|F=12)=1$

Step 1:Given Information

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.

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

Step 2:Table Representation

The sum of numbers of two dice are,

Sum	Possible outcomes	Probability
$2$	$(1,1)$	$1/36$
$3$	$(1,2),(2,1)$	$2/36$
$4$	$(1,3),(2,2),(3,1)$	$3/36$
$5$	$(1,4),(2,3),(3,2),(4,1)$	4/36
$6$	$(1,5),(2,4),(3,3),(4,2),(5,1)$	5/36
$7$	$(1,6),(2,5),(3,4),(4,3),(5,2),(6,1)$	6/36
$8$	$(2,6),(3,5),(4,4),(5,3),(6,2)$	5/36
$9$	$(3,6),(4,5),(5,4),(6,3)$	4/36
$10$	$(4,6),(5,5),(6,4)$	3/36
$11$	$(5,6),(6,5)$	$2/36$
$12$	$(6,6)$	$1/36$
Total	$36$	1

Step 3:Explanation

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,

$P(E\mid F=2)=0$
$P(E\mid F=3)=0$

$P(E\mid F=4)=0$

$P(E\mid F=5)=0$

$P(E\mid F=6)=0$

Step 4:Conditional Probability of ( E / F = 7 )

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:

$E\cap F=\left\{\right(1,6),(6,1\left)\right\}$

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,

$P(E\cap F=7)=\frac{2}{36}$

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,

$P(E\mid F=7)=\frac{P(E\cap F=7)}{P(F=7)}$

$=\frac{{\displaystyle \frac{2}{36}}}{{\displaystyle \frac{6}{36}}}$

$=\frac{2}{6}$

Step 5:Conditional Probability of ( E | F = 8 )

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:

$E\cap F=\left\{\right(2,6),(6,2\left)\right\}$

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,

$P(E\cap F=8)=\frac{2}{36}$

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,

$P(E\mid F=8)=\frac{P(E\cap F=8)}{P(F=8)}$

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

$=\frac{2}{5}$

Step 6:Conditional Probability of ( E / F = 9 )

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:

$E\cap F=\left\{\right(3,6),(6,3\left)\right\}$

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$9$,

$P(E\cap F=9)=\frac{2}{36}$

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,

$P(E\mid F=9)=\frac{P(E\cap F=9)}{P(F=9)}$

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

$=\frac{2}{4}$

Step 7:Conditional Probability of ( E / F = 10 )

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:

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

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$10$,

$P(E\cap F=10)=\frac{2}{36}$

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,

$P(E\mid F=10)=\frac{P(E\cap F=10)}{P(F=10)}$

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

$=\frac{2}{3}$

Step 8:Conditional probability of ( E / F = 11 )

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:

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

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,

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

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,

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

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

$=1$

Step 9:Conditional Probability of ( E / F = 12 )

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:

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

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,

$P(E\cap F=12)=\frac{1}{36}$

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,

$P(E\mid F=12)=\frac{P(E\cap F=12)}{P(F=12)}$

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

$=1$

Step 10:Final Answer

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" style="max-width: none; vertical-align: -4px;" $1$

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