JEE Advance - Mathematics (2021 - Paper 2 Online - No. 17)
A number of chosen at random from the set {1, 2, 3, ....., 2000}. Let p be the probability that the chosen number is a multiple of 3 or a multiple of 7. Then the value of 500p is __________.
Answer
214
Explanation
Given, set = {1, 2, 3, ...., 2000}
Let E1 = Event that it is a multiple of 3 = {3, 6, 9, ...., 1998}
$$\therefore$$ n(E1) = 666
and E2 = Event that it is a multiple of 7 = {7, 14, ..., 1995}
$$\therefore$$ n(E2) = 285
$${E_1} \cap {E_2}$$ = multiple of 21 = {21, 42, ....., 1995}
$$n({E_1} \cap {E_2}) = 95$$
$$\therefore$$ $$P({E_1} \cup {E_2}) = P({E_1}) + P({E_2}) - P({E_1} \cap {E_2})$$
$$P({E_1} \cup {E_2}) = {{666 + 285 - 95} \over {2000}} = {{856} \over {2000}} = p$$ (given)
Hence, $$500p = 500 \times {{856} \over {2000}} = {{856} \over 4} = 214$$
Let E1 = Event that it is a multiple of 3 = {3, 6, 9, ...., 1998}
$$\therefore$$ n(E1) = 666
and E2 = Event that it is a multiple of 7 = {7, 14, ..., 1995}
$$\therefore$$ n(E2) = 285
$${E_1} \cap {E_2}$$ = multiple of 21 = {21, 42, ....., 1995}
$$n({E_1} \cap {E_2}) = 95$$
$$\therefore$$ $$P({E_1} \cup {E_2}) = P({E_1}) + P({E_2}) - P({E_1} \cap {E_2})$$
$$P({E_1} \cup {E_2}) = {{666 + 285 - 95} \over {2000}} = {{856} \over {2000}} = p$$ (given)
Hence, $$500p = 500 \times {{856} \over {2000}} = {{856} \over 4} = 214$$
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