JEE MAIN - Mathematics (2008 - No. 6)
A die is thrown. Let $$A$$ be the event that the number obtained is greater than $$3.$$ Let $$B$$ be the event that the number obtained is less than $$5.$$ Then $$P\left( {A \cup B} \right)$$ is :
$${3 \over 5}$$
$$0$$
$$1$$
$${2 \over 5}$$
Explanation
No of outcome for a die = { 1, 2, 3, 4, 5, 6 }
According to the question,
A = { 4, 5, 6 }
$$\therefore$$ P(A) = $${3 \over 6}$$
B = { 1, 2, 3, 4 }
$$\therefore$$ P(A) = $${4 \over 6}$$
A $$ \cap $$ B = { 4 }
So P(A $$ \cap $$ B) = $${1 \over 6}$$
We know, $$P\left( {A \cup B} \right)$$ = $$P\left( A \right)$$ + $$P\left( B \right)$$ - $$P\left( {A \cap B} \right)$$
$$\therefore$$ $$P\left( {A \cup B} \right)$$ = $${3 \over 6}$$ + $${4 \over 6}$$ - $${1 \over 6}$$ = $${{7 - 1} \over 6}$$ = $${6 \over 6}$$ = 1
$$\therefore$$ Option (C) is correct.
According to the question,
A = { 4, 5, 6 }
$$\therefore$$ P(A) = $${3 \over 6}$$
B = { 1, 2, 3, 4 }
$$\therefore$$ P(A) = $${4 \over 6}$$
A $$ \cap $$ B = { 4 }
So P(A $$ \cap $$ B) = $${1 \over 6}$$
We know, $$P\left( {A \cup B} \right)$$ = $$P\left( A \right)$$ + $$P\left( B \right)$$ - $$P\left( {A \cap B} \right)$$
$$\therefore$$ $$P\left( {A \cup B} \right)$$ = $${3 \over 6}$$ + $${4 \over 6}$$ - $${1 \over 6}$$ = $${{7 - 1} \over 6}$$ = $${6 \over 6}$$ = 1
$$\therefore$$ Option (C) is correct.
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