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In the random experiment of tossing two unbiased dice let $E$ be the event of getting the sum 8 and $F$ be the event of getting even numbers on both the dice. Then :
I. $P(E)=\frac{7}{36}$
II. $P(F)=\frac{1}{3}$
Which of the following is a correct statement?
Options:
I. $P(E)=\frac{7}{36}$
II. $P(F)=\frac{1}{3}$
Which of the following is a correct statement?
Solution:
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Verified Answer
The correct answer is:
Neither I nor II is true
$E=$ Event of getting the sum 8 .
$=\{(2,6),(3,5),(4,4),(5,3),(6,2)\}$
and $F=$ Event of getting the even numbers on both the dice.
$=\{(1,1),(2,2),(3,3),(4,4),(5,5),(6,6)\}$
$\therefore \quad P(E)=5 \quad$ and $\quad P(F)=6$
also $\quad n(S)=36$
$\therefore \quad P(E)=\frac{n(E)}{n(S)}=\frac{5}{36}$
and $\quad P(F)=\frac{n(F)}{n(S)}=\frac{6}{36}$
$=\frac{1}{6}$
$\therefore \quad$ Neither I nor II is true
$=\{(2,6),(3,5),(4,4),(5,3),(6,2)\}$
and $F=$ Event of getting the even numbers on both the dice.
$=\{(1,1),(2,2),(3,3),(4,4),(5,5),(6,6)\}$
$\therefore \quad P(E)=5 \quad$ and $\quad P(F)=6$
also $\quad n(S)=36$
$\therefore \quad P(E)=\frac{n(E)}{n(S)}=\frac{5}{36}$
and $\quad P(F)=\frac{n(F)}{n(S)}=\frac{6}{36}$
$=\frac{1}{6}$
$\therefore \quad$ Neither I nor II is true
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