In a single throw of two dice, find the probability of
(i) getting a sum less than 6
(ii) getting a doublet of odd numbers
(iii) getting the sum as a prime number
(i) We know that,
Probability of occurrence of an event
$=\frac{\text { Total no. of Desired outcomes }}{\text { Total no.of outcomes }}$
Outcomes are $(1,1),(1,2),(1,3),(1,4),(1,5),(1,6)$,
$(2,1),(2,2),(2,3),(2,4),(2,5),(2,6)$,
$(3,1),(3,2),(3,3),(3,4),(3,5),(3,6)$,
$(4,1),(4,2),(4,3),(4,4),(4,5),(4,6)$,
$(5,1),(5,2),(5,3),(5,4),(5,5),(5,6)$,
$(6,1),(6,2),(6,3),(6,4),(6,5),(6,6)$
Total no.of outcomes are 36
In that only $(1,1),(1,2),(1,3),(1,4),(2,1),(2,2),(2,3),(3,1),(3,2),(4,1)$ are our desired outputs as there sum is less than 6
Therefore no.of desired outcomes are 10
Therefore, the probability of getting a sum less than 6
$=\frac{10}{36}$
$=\frac{5}{18}$
Conclusion: Probability of getting a sum less than 6, when two dice are rolled is $\frac{5}{18}$
(ii) We know that,
Probability of occurrence of an event
$=\frac{\text { Total no. of Desired outcomes }}{\text { Total no. of outcomes }}$
In $(a, b)$ if $a=b$ then it is called a doublet
Total doublets are $(1,1),(2,2),(3,3),(4,4),(5,5),(6,6)$
In $(a, b)$ if $a=b$ and if $a, b$ both are odd then it is called a doublet
Odd doublets are $(1,1),(3,3),(5,5)$
Outcomes are $(1,1),(1,2),(1,3),(1,4),(1,5),(1,6)$,
$(2,1),(2,2),(2,3),(2,4),(2,5),(2,6)$,
$(3,1),(3,2),(3,3),(3,4),(3,5),(3,6)$,
$(4,1),(4,2),(4,3),(4,4),(4,5),(4,6)$,
$(5,1),(5,2),(5,3),(5,4),(5,5),(5,6)$,
$(6,1),(6,2),(6,3),(6,4),(6,5),(6,6)$
Total no. of outcomes are 36 and desired outcomes are 3
Therefore, probability of getting doublet of odd numbers
$=\frac{3}{36}$
$=\frac{1}{12}$
Conclusion: Probability of getting doublet of odd numbers, when two dice are rolled
is $\frac{1}{12}$
(iii) We know that,
Probability of occurrence of an event
$=\frac{\text { Total no.of Desired outcomes }}{\text { Total no. of outcomes }}$
Outcomes are $(1,1),(1,2),(1,3),(1,4),(1,5),(1,6)$,
$(2,1),(2,2),(2,3),(2,4),(2,5),(2,6)$,
$(3,1),(3,2),(3,3),(3,4),(3,5),(3,6)$,
$(4,1),(4,2),(4,3),(4,4),(4,5),(4,6)$,
$(5,1),(5,2),(5,3),(5,4),(5,5),(5,6)$,
$(6,1),(6,2),(6,3),(6,4),(6,5),(6,6)$
Total no. of outcomes are 36
Desired outputs are $(1,1),(1,2),(1,4),(1,6),(2,1),(2,3),(2,5),(3,2),(3,4),(4,1)$, $(4,3),(5,2),(5,6),(6,1),(6,5)$
Total no. of desired outputs are 15
Therefore, probability of getting the sum as a prime number
$=\frac{15}{36}$
$=\frac{5}{12}$
Conclusion: Probability of getting the sum as a prime number, when two dice are rolled
is $\frac{5}{12}$
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