Question

You roll two fair dice, a green one and a red one.

(a) What is the probability of getting a sum of 7? (Enter your answer as a fraction.)
1

(b) What is the probability of getting a sum of 11? (Enter your answer as a fraction.)
2

(c) What is the probability of getting a sum of 7 or 11? (Enter your answer as a fraction.)


Are these outcomes mutually exclusive?

Yes No    

Answer 1

Concepts and reason

The concept of probability is used to solve this question.

Probability is defined as an extent to which an event can occur. The probability of an event always lies between 0 and 1.

The probability will be calculated by generating the favorable outcomes and sample space of an event.

Fundamentals

Probability is defined as the number of outcomes generated from an event divided by the sample space. Sample space is defined as the total outcomes of an event.

$P = \frac{{{\rm{Number of outcomes}}}}{{{\rm{Sample space}}}}$

Consider that two events, A and B, occur with the probabilities $P\left( A \right){\rm{ and }}P\left( B \right)$ respectively. The probability that both the events occur is,

$P\left( {A{\rm{ and }}B} \right) = P\left( {A \cap B} \right)$

When both the events are independent,

$\begin{array}{c}\\P\left( {A{\rm{ and }}B} \right) = P\left( {A \cap B} \right)\\\\ = P\left( A \right) \times P\left( B \right)\\\end{array}$

The probability that either A or B occurs is,

$\begin{array}{c}\\P\left( {A{\rm{ or B}}} \right) = P\left( {A \cup B} \right)\\\\ = P\left( A \right) + P\left( B \right) - P\left( {A \cap B} \right)\\\end{array}$

The probability of an event always lies between 0 and 1.

(a)

When two fair dice are rolled, one green and the other red, then the total number of outcomes (sample space) is 36.

The sample space is as follows,

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

The probability of getting the sum 11 is calculated as,

$\begin{array}{c}\\{\rm{Number of outcomes having sum of 7}} = \left\{ {\left( {{\rm{1,6}}} \right)\left( {{\rm{2,5}}} \right)\left( {{\rm{3,4}}} \right)\left( {{\rm{4,3}}} \right)\left( {{\rm{5,2}}} \right)\left( {{\rm{6,1}}} \right)} \right\}\\\\P\left( {{\rm{the sum of two dices are 7}}} \right) = \frac{{{\rm{Number of outcomes}}}}{{{\rm{Sample space}}}}\\\\ = \frac{{\rm{6}}}{{{\rm{36}}}}\\\\ = {\rm{0}}{\rm{.1666}}\\\end{array}$

(b)

The probability of getting sum of 11 when two dices are rolled is calculated as,

$\begin{array}{c}\\{\rm{Number of outcomes having sum of}}11 = \left\{ {\left( {6,5} \right)\left( {5,6} \right)} \right\}\\\\P\left( {{\rm{the sum of two dices are }}11} \right) = \frac{{{\rm{Number of outcomes}}}}{{{\rm{Sample space}}}}\\\\ = \frac{{\rm{2}}}{{36}}\\\\ = 0.055\\\end{array}$

(c)

The probability of getting a sum of 7 or 11 is calculated as,

The probability of getting the sum of 11 is calculated as,

$\begin{array}{c}\\{\rm{Number of outcomes having sum of}}11 = \left\{ {\left( {6,5} \right)\left( {5,6} \right)} \right\}\\\\P\left( {{\rm{the sum of two dices are }}11} \right) = \frac{{{\rm{Number of outcomes}}}}{{{\rm{Sample space}}}}\\\\ = \frac{{\rm{2}}}{{36}}\\\\ = {\rm{0}}{\rm{.055}}\\\end{array}$

The probability of getting the sum of 7 is calculated as,

$\begin{array}{c}\\{\rm{Number of outcomes having sum of }}7 = \left\{ {\left( {1,6} \right)\left( {6,1} \right)\left( {2,5} \right)\left( {5,2} \right)\left( {3,4} \right)\left( {4,3} \right)} \right\}\\\\P\left( {{\rm{the sum of two dices are }}7} \right) = \frac{{{\rm{Numberofoutcomes}}}}{{{\rm{Samplespace}}}}\\\\ = \frac{6}{{36}}\\\\ = {\rm{0}}{\rm{.1666}}\\\end{array}$

The combined probability of getting 7 or 11 is calculated as,

$\begin{array}{c}\\P\left( {{\rm{7or11}}} \right) = P\left( {{\rm{getting }}7} \right) + P\left( {{\rm{getting }}11} \right) - P\left( {{\rm{getting 7 and 11}}} \right)\\\\ = {\rm{0}}{\rm{.055}} + 0.1666 + 0\\\\ = {\rm{0}}{\rm{.2216}}\\\end{array}$

Ans: Part a

The probability of two dices summing to 7 is $0.1666$ .