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Two different dice are thrown together. Find the probability that the numbers obtained have

(i) even sum, and

(ii) even product.

Answers (1)

Given $\rightarrow$ Two different dice are thrown.

Total number of outcomes when two dice are thrown are n(T)

Total outcomes $\rightarrow$

$\left \{ (1,1),(1,2),(1,3),(1,4),(1,5),(1,6),$

$(2,1),(2,2),(2,3),(2,4),(2,5),(2,6),$

$(3,1),(3,2),(3,3),(3,4),(3,5),(3,6),$

$(4,1),(4,2),(4,3),(4,4),(4,5),(4,6),$

$(5,1),(5,2),(5,3),(5,4),(5,5),(5,6),$

$(6,1),(6,2),(6,3),(6,4),(6,5),(6,6) \}$

$\therefore n(T)=36$

(i) Probablity of finding even sum $=P_1$

Outcomes when sum is even

$=(1,1),(1,3),(1,5),(2,2),(2,4),(2,6),$

$(3,1),(3,3),(3,5),(4,2),(4,4),(4,6),$

$(5,1),(5,3),(5,5),(6,2),(6,4),(6,6)$

$P_1=\frac{n(P_1)}{n(T)}=\frac{18}{36}=\frac{1}{2}$

(ii) Probablity of finding even product $=P_2$

Outcomes when product is even

$n(P_2)=(1,2),(1,4),(1,6),(2,1),(2,2),(2,3),(2,4),(2,5),(2,6),(3,2),(3,4),(3,6),(4,1),(4,2),$ $(4,3),(4,4),(4,5),(4,6),(5,2),(5,4),(5,6),(6,1),(6,2),(6,3),(6,4),(6,5),(6,6)$

$n(P_2)=27$

$P_2=\frac{n(P_2)}{n(T)}=\frac{27}{36}=\frac{3}{4}$

Posted by

Ravindra Pindel

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