# Two dice are thrown together. What is the probability that the sum of the numbers on two dice is \[5\] or number on the second die is greater than or equal to the number on the first die?

Answer

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Hint- To determine the probability, first we have to define the event and make the sample space and then proceed further using probability definition.

Given that the two dice are thrown together. So the total possible conditions will be

$6 \times 6 = 36$

Let ${E_1}$ be the event of getting sum as $5$

The possible combination of the top faces of the die for the event ${E_1}$

${E_1} = \{ (1,4),(4,1),(2,3),(3,2)\} $

The probability of event \[\;\;{E_1} = P({E_1}) = \dfrac{4}{{36}}\]

Let ${E_2}$ be the event of getting a number greater or equal on the top face of the second die than that of the first die.

The possible combinations of top face of die for event ${E_2}$

$

{E_2} = \{ (1,1),(1,2),(1,3),(1,4),(1,5),(1,6),(2,2),(2,3),(2,4),(2,5),(2,6), \\

(3,3),(3,4),(3,5),(3,6), \\

(4,4),(4,5),(4,6), \\

(5,5),(5,6), \\

(6,6)\} \\

$

The probability of the event ${E_2} = P({E_2}) = \dfrac{{21}}{{36}}$

Let $E$ be the event that one of the two events ${E_1}$ or ${E_2}$ will occur

Hence the probability of event $E$ will be sum of the probability of events ${E_1}$and ${E_2}$

$

P(E) = P({E_1}) + P({E_2}) \\

P(E) = \dfrac{4}{{36}} + \dfrac{{21}}{{36}} = \dfrac{{25}}{{36}} \\

$

Note- In this type of numerical first try to declare the events and make the sample space and keep in mind the conditions given in the question and proceed according to the conditions. Probability of occurrence of an event is ratio of number of favorable outcomes over total number of possible outcomes of the given event.

Given that the two dice are thrown together. So the total possible conditions will be

$6 \times 6 = 36$

Let ${E_1}$ be the event of getting sum as $5$

The possible combination of the top faces of the die for the event ${E_1}$

${E_1} = \{ (1,4),(4,1),(2,3),(3,2)\} $

The probability of event \[\;\;{E_1} = P({E_1}) = \dfrac{4}{{36}}\]

Let ${E_2}$ be the event of getting a number greater or equal on the top face of the second die than that of the first die.

The possible combinations of top face of die for event ${E_2}$

$

{E_2} = \{ (1,1),(1,2),(1,3),(1,4),(1,5),(1,6),(2,2),(2,3),(2,4),(2,5),(2,6), \\

(3,3),(3,4),(3,5),(3,6), \\

(4,4),(4,5),(4,6), \\

(5,5),(5,6), \\

(6,6)\} \\

$

The probability of the event ${E_2} = P({E_2}) = \dfrac{{21}}{{36}}$

Let $E$ be the event that one of the two events ${E_1}$ or ${E_2}$ will occur

Hence the probability of event $E$ will be sum of the probability of events ${E_1}$and ${E_2}$

$

P(E) = P({E_1}) + P({E_2}) \\

P(E) = \dfrac{4}{{36}} + \dfrac{{21}}{{36}} = \dfrac{{25}}{{36}} \\

$

Note- In this type of numerical first try to declare the events and make the sample space and keep in mind the conditions given in the question and proceed according to the conditions. Probability of occurrence of an event is ratio of number of favorable outcomes over total number of possible outcomes of the given event.

Last updated date: 25th Sep 2023

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