Q. 6.1 - A First Course in Probability

Question

Each throw of an unfair die lands on each of the odd numbers $1, 3, 5$ with probability C and on each of the even numbers with probability $2C$.

 (a) Find C.

 (b) Suppose that the die is tossed. Let X equal $1$ if the result is an even number, and let it be $0$ otherwise. Also, let Y equal $1$ if the result is a number greater than three and let it be $0$ otherwise. Find the joint probability mass function of X and Y. Suppose now that $12$ independent tosses of the die are made.

 (c) Find the probability that each of the six outcomes occurs exactly twice.

 (d) Find the probability that $4$ of the outcomes are either one or two, $4$ are either three or four, and $4$ are either five or six.

 (e) Find the probability that at least $8$ of the tosses land on even numbers. 

Step-by-Step Solution

Answer

$\left(a\right) C=\frac{1}{9}$

$$\begin{gathered} \left(b\right) P(X=0,Y=0)=\frac{2}{9} \\ P(X=0,Y=1)=\frac{1}{9} \\ P(X=1,Y=0)=\frac{2}{9} \\ P(X=1,Y=1)=\frac{4}{9} \end{gathered}$$

(c) the probability that each of the six outcomes occurs exactly twice $=\frac{12!}{2^6}{\left(\frac{1}{3}\right)}^6{\left(\frac{2}{3}\right)}^6$

(d) the probability that at least 8 of the tosses land on even numbers $=\frac{12!}{4!4!4!}{\left(\frac{1}{3}\right)}^{12}$

(e) the probability that at least 8 of the tosses land on even numbers$=\sum _{k=8}^{12}\frac{12!}{k!(12-k)!}{\left(\frac{2}{3}\right)}^k{\left(\frac{1}{3}\right)}^{12-k}$

Step 1: Given information (part a)

C is a probability that an unfair die lands on each of the odd numbers $1,3,5$ and $2C$ is a probability that an unfair die lands on each of the even numbers.

Step 2: Explanation (part a)

The sum of probabilities must be equal to $1$.

$$\begin{gathered} 3C+3\left(2C\right)=1 \\ \Rightarrow 3C+6C=1 \\ \Rightarrow 9C=1 \\ \Rightarrow C=\frac{1}{9} \end{gathered}$$

Step 3: Given information (part b)

If the result is an even number then X is equal to $1$. Otherwise, let $X=0$.

Step 4: Explanation (part b)

The joint probability mass function of two discrete random variables X and Y is equal to $P_{XY}\left(x,y\right)=P\left(X=x, Y=y\right)$

From part a.,

Probability that an unfair die lands on the odd number$=C=\frac{1}{9}$

Probability that an unfair die lands on an even number$=2C=\frac{2}{9}$

Case $1: X=0, Y=0$

$X=0$means number can be  $1,3 or 5$

$Y=0$means number can be $1,2 or 3$

So, $X=0$, $Y=0$ means number can be $1 or 3$

Also,

Each throw of an unfair die lands on each of the odd numbers $1,3,5$ with probability $C=C=\frac{1}{9}$

Therefore,

$P\left(X=0, Y=1\right)$ is equivalent to finding P (number is $5$)$=\frac{1}{9}$

So,

$P\left(X=0, Y=1\right)=\frac{1}{9}$

Case $3:X=1, Y=0$

$X=1$ means number can be $2,4 or 6$

$Y=0$ means number can be $1,2 or 3$

So, $X=1, Y=0$means number can be only $2$

Also,

Each throw of an unfair die lands on each of the even numbers $2,4,6$ with probability$=2C=\frac{2}{9}$

Therefore,

$P\left(X=1, Y=0\right)$ is equivalent to finding P (number is $2$)

$P\left(X=1, Y=0\right)=P\left(number is 2\right)=\frac{2}{9}$

So, 

$P\left(X=1, Y=0\right)=\frac{2}{9}$

$Case 4:X=1, Y=1$

$X=1$$means can be 2,4 or 6$

$Y=1 means number can be 4,5 or 6$

So, $X=1, Y=1 means number can be only 4 or 6$

Each throw of an unfair die lands on each of the even numbers $2,4,6 with probability C=2C=\frac{2}{9}$

Therefore,

$P\left(X=1, Y=1\right) is equivalent to finding P\left(number on die is 4 or 6\right)$

As events: number on die is $4$ and number on die is $6$ are disjoint

$$\begin{gathered} P\left(X=1, Y=1\right)=P\left(X=1\right)+P\left(Y=1\right)=\frac{2}{9}+\frac{2}{9}=\frac{4}{9} \\ \left\{\because P\left(A\cup B\right)=P\left(A\right)+P\left(B\right) if A\cap B=\varnothing \right\} \end{gathered}$$

Step 5: Given information (part c)

$12$ independent tosses of a die are made

Formula used: $Probability =\frac{Number of favourable outcomes}{Total number of outcomes}$

Step 6: Explanation (part c)

Probability that the number on the die is odd$=C=\frac{1}{9}$

Probability that the number on the die is even$=2C=\frac{2}{9}$

Total of ways of getting two ones, two twos, two threes, two fours, two fives and two sixes$=\frac{12!}{(2!{)}^6}=\frac{12!}{2^6}$

(as Total number of terms is $12$ and there is a repetition of two ones, two twos, two threes, two fours, two fives, two sixes)

Probability of each of the ways$={\left[3\left(\frac{1}{9}\right)\right]}^6{\left[3\left(\frac{2}{9}\right)\right]}^6={\left(\frac{1}{3}\right)}^6{\left(\frac{2}{3}\right)}^6$

(as three numbers are even and three are odd)

So,

Probability that each of the six outcomes occurs exactly twice$=\frac{12!}{2^6}{\left(\frac{1}{3}\right)}^6{\left(\frac{2}{3}\right)}^6$

Step 7: Given information (part d)

Probability that the number on the die is odd$=C=\frac{1}{9}$

Probability that the number on the die is even $=2C=\frac{2}{9}$

Events: getting even number or odd number are disjoint events.

Step 8: Explanation (part d)

P(number is even or odd) = (number is even) + P(number is odd)

$=\frac{1}{9}+\frac{2}{9}=\frac{3}{9}=\frac{1}{3}$

So, each of the events: number is one or two, number is three or four, number is five or six has he same probability $\frac{1}{3}$

Now, if $4$ out of the comes are either one or two, $4$ out of the outcomes three or four, $4$ of the outcomes are either five or six,

Total number of ways$=\frac{12!}{4!4!4!}$

(as number of terms is $12$ and there are four repetitions of each of the three numbers)

Each of these ways has the same probability to occur equal to $={\left(\frac{1}{3^{}}\right)}^{12}$

Therefore,

Probability that $4$ out of the outcomes are either one or two, 4 are either three or four, $4$ are either five or six $=\frac{12!}{4!4!4!}{\left(\frac{1}{3}\right)}^{12}$

Step 9: Given information (part e)

The binomial probability formula is $P\left(X\right)=b(n,p)=\frac{n!}{(n-x)!x!}{p}^x(1-p{)}^x$

her p denotes the probability of success.

Step 10: Explanation (part e)

let x denotes a random variable that marks the number of times he throws of an unfair die lands on the even number.

p = $P\left(\text{ the throw of an unfair die lands on the even number }\right)=3\left(\frac{2}{9}\right)=\frac{2}{3}$

so $X~b\left(12,\frac{2}{3}\right)$

therefore

the probability that at least 8 of the tosses land on even numbers$=P\left(X\ge 8\right)$

$=\sum _{k=8}^{12}\frac{12!}{k!(12-k)!}{\left(\frac{2}{3}\right)}^k{\left(\frac{1}{3}\right)}^{12-k}$