Question

1. You roll a pair of standard dice. Create the sample space for a single roll of the dice and use the sample space to compute the following probabilities.

A. Create a sample space.

B. P ( getting a 1 on the first die or getting a 5 on the second die)

C. P (Sum of the dice = 10)

D. P ( getting a 3 on the second die given that you got a 2 on the first die)

E. P (getting an odd number on the first die and a value greater than 4 on the second die)

Answer 1

Concepts and reason

Event:

The collection or the set of outcomes in an experiment is called as an event.

Sample space:

All the possible outcomes in an experiment are called as sample space.

Probability:

The ratio of the number of favorable outcomes to certain event and total number of possible outcomes is called as the probability of an event.

Mutually Exclusive events:

Two events are said to be mutually exclusive if they cannot occur at the same time. It is also known as disjoint event.

Non-mutually exclusive events:

Two events are said to be non-mutually exclusive if they occur or happened at the same time.

Intersection of two events:

The set of the outcomes that belong to both the two events is called as intersection of two events.

Conditional probability:

The probability of happening of an event given that another event has already happened is called as the conditional probability.

Fundamentals

The probability of an event is defined as,

Addition Rule of Probability:

Let A and B be two events, then the probability for event A or event B can be defined as,

$P\left( {A \cup B} \right) = P\left( A \right) + P\left( B \right) - P\left( {A \cap B} \right)$

If the events A and B are mutually exclusive or disjoint then $P\left( {A \cap B} \right) = 0$ .Thus, the addition rule for mutually exclusive event is defined as,

$P\left( {A\,{\rm{or}}\,B} \right) = P\left( A \right) + P\left( B \right)$

Conditional probability:

Let A and B be two events, then the conditional probability for event A, given that event B has already happened, can be defined as,

Also, $P\left( {A \cap B} \right)$ is the probability of the common outcomes of both the events A and B.

Formula to find the probability of event $\left( {A \cap B} \right)$ is $P\left( {A \cap B} \right) = \frac{{N\left( {A \cap B} \right)}}{{N\left( S \right)}}$ .

(1.A)

The sample space is obtained as shown below:

From the information, a rolling die roll once. The sample space of rolling a two die is given.

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

(1.B)

The probability that the getting a 1 on the first die or getting a 5 on the second die is obtained as shown below:

Let the event A denotes the 1 on the first die and event B denotes 5 on the second die. From the information, the sample space of rolling a two die is given.

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

The probability that the getting a 1 on the first die or getting a 5 on the second die is,

$\begin{array}{c}\\P\left( {A \cup B} \right) = P\left( A \right) + P\left( B \right) - P\left( {A \cap B} \right)\\\\ = \frac{6}{{36}} + \frac{6}{{36}} - \frac{1}{{36}}\\\\ = \frac{{6 + 6 - 1}}{{36}}\\\\ = 0.3056\\\end{array}$

(1.C)

The probability that the sum of the dice is 10 is obtained as shown below:

Let the event C denotes the sum of the dice is 10.

From the information, the sample space of rolling a two die is given.

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

The number of outcomes that the sum of the dice is 10 are $\left( {4,\,6} \right)$ , $\left( {5,\,5} \right)$ and $\left( {6,\,4} \right)$ . That is, $N\left( C \right) = 3$ .

The probability that the sum of the dice is 10 is,

$\begin{array}{c}\\P\left( C \right) = \frac{{N\left( C \right)}}{{N\left( S \right)}}\\\\ = \frac{3}{{36}}\\\\ = 0.0833\\\end{array}$

(1.D)

The probability that the getting a 3 on the second die given that you got a 2 on the first die is obtained as shown below:

Let the event D denotes the 2 on the first die and event E denotes 3 on the second die.

From the information, the sample space of rolling a two die is given.

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

The number of outcomes of 2 on the first die are $\left( {2,\,1} \right),\,\left( {2,\,2} \right),\,\left( {2,\,3} \right),\left( {2,\,4} \right),\,\left( {2,\,5} \right)\,{\rm{and}}\,\left( {2,\,6} \right)$ . The number of outcomes of 3 on the second die are $\left( {1,\,3} \right)$ , $\left( {2,\,3} \right)$ , $\left( {3,\,3} \right)$ , $\left( {4,\,3} \right)$ , $\left( {5,\,3} \right)$ and $\left( {6,\,3} \right)$ . That is, $N\left( D \right) = 6$ , $N\left( E \right) = 6$ and $N\left( {E \cap D} \right) = 1$ .

The probability that the getting a 3 on the second die given that you got a 2 on the first die is,

$\begin{array}{c}\\P\left( {E|D} \right) = \frac{{P\left( {E \cap D} \right)}}{{P\left( D \right)}}\\\\ = \frac{{\left( {\frac{1}{{36}}} \right)}}{{\left( {\frac{6}{{36}}} \right)}}\\\\ = \frac{1}{6}\\\\ = 0.1667\\\end{array}$

(1.E)

The probability that the getting an odd number on the first die and a value greater than 4 on the second die is obtained as shown below:

Let the event F denotes an odd number on the first die and event G denotes a value greater than 4 on the second die.

From the information, the sample space of rolling a two die is given.

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

The number of outcomes of an odd number on the first die are,

$\left\{ \begin{array}{l}\\\left( {1,\,1} \right),\,\left( {1,\,2} \right),\,\left( {1,\,3} \right),\left( {1,\,4} \right),\,\left( {1,\,5} \right),\,\left( {1,\,6} \right),\\\\\left( {3,\,1} \right),\,\left( {3,\,2} \right),\,\left( {3,\,3} \right),\left( {3,\,4} \right),\,\left( {3,\,5} \right),\,\left( {3,\,6} \right),\\\\\left( {5,\,1} \right),\,\left( {5,\,2} \right),\,\left( {5,\,3} \right),\left( {5,\,4} \right),\,\left( {5,\,5} \right)\,{\rm{and}}\,\left( {5,\,6} \right)\\\end{array} \right\}$ .

The number of outcomes of a value greater than 4 on the second die are

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

That is, $N\left( F \right) = 18$ , $N\left( G \right) = 12$ and $N\left( {F \cap G} \right) = 6$ .

The probability that the getting an odd number on the first die and a value greater than 4 on the second die is,

$\begin{array}{c}\\P\left( {F \cap G} \right) = P\left( {F \cap G} \right)\\\\ = \frac{{N\left( {F \cap G} \right)}}{{N\left( S \right)}}\\\\ = \frac{6}{{36}}\\\\ = 0.1667\\\end{array}$

Ans: Part 1.A

The sample space is,

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

Part 1.B

The probability that the getting a 1 on the first die or getting a 5 on the second die is 0.3056.

Part 1.C

Thus, the probability that the sum of the dice is 10 is 0.0833.

Part 1.D

The probability that the getting a 3 on the second die given that you got a 2 on the first die is 0.1667.

Part 1.E

The probability that the getting an odd number on the first die and a value greater than 4 on the second die is 0.1667.