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1

### JEE Advanced 2021 Paper 2 Online

Numerical
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 __________.

## 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, $$500p = 500 \times {{856} \over {2000}} = {{856} \over 4} = 214$$
2

### JEE Advanced 2021 Paper 1 Online

Numerical
Three numbers are chosen at random, one after another with replacement, from the set S = {1, 2, 3, ......, 100}. Let p1 be the probability that the maximum of chosen numbers is at least 81 and p2 be the probability that the minimum of chosen numbers is at most 40.

The value of $${{125} \over 4}{p_2}$$ is ___________.

## Explanation

p2 = 1 $$-$$ p (all three numbers are > 40)

$$= 1 - {\left( {{{60} \over {100}}} \right)^3} = 1 - {{27} \over {125}} = {{98} \over {125}}$$

So, $${{125{p_2}} \over 4} = {{125} \over 4} \times {{98} \over {125}} = {{98} \over 4} = 24.50$$
3

### JEE Advanced 2021 Paper 1 Online

Numerical
Three numbers are chosen at random, one after another with replacement, from the set S = {1, 2, 3, ......, 100}. Let p1 be the probability that the maximum of chosen numbers is at least 81 and p2 be the probability that the minimum of chosen numbers is at most 40.

The value of $${{625} \over 4}{p_1}$$ is ___________.

## Explanation

p1 = 1 $$-$$ p (all 3 numbers are $$\le$$ 80)

$$= 1 - {\left( {{{80} \over {100}}} \right)^3} = {{125 - 64} \over {125}} = {{61} \over {125}}$$

So, $${{625{p_1}} \over 4} = {{625} \over 4} \times {{61} \over {125}} = {5 \over 4} \times 61 = 76.25$$
4

### JEE Advanced 2020 Paper 2 Offline

Numerical
Two fair dice, each with faces numbered 1, 2, 3, 4, 5 and 6, are rolled together and the sum of the numbers on the faces is observed. This process is repeated till the sum is either a prime number or a perfect square. Suppose the sum turns out to be a perfect square before it turns out to be a prime number. If p is the probability that this perfect square is an odd number, then the value of 14p is ..........

## Explanation

Let an event E of sum of outputs are perfect square (i.e., 4 or 9), so

E = {(1, 3), (2, 2), (3, 1), (3, 6), (4, 5), (5, 4), (6, 3)}

and an event F of sum of outputs are prime numbers

(i.e., 2, 3, 5, 7, 11) so

F = {(1, 1), (1, 2), (2, 1), (1, 4), (2, 3), (3, 2), (4, 1), (1, 6), (2, 5), (3, 4), (4, 3), (5, 2), (6, 1), (5, 6), (6, 5)}

and event T of sum of outputs are odd numbers

(i.e., 3, 5, 7, 9, 11)

T = {(1, 2), (2, 1), (1, 4), (2, 3), (3, 2), (4, 1), (1, 6), (2, 5), (3, 4), (4, 3), (5, 2), (6, 1), (3, 6), (4, 5), (5, 4), (6, 3), (5, 6), (6, 5)}

Now, required probability $$p = P(T/E)$$

$$= {{P(T \cap E)} \over {P(E)}}$$

where, $${P(T \cap E)}$$ = probability of occurring perfect square odd number before prime

$$= \left( {{4 \over {36}}} \right) + \left( {{{14} \over {36}}} \right)\left( {{4 \over {36}}} \right) + {\left( {{{14} \over {36}}} \right)^2}\left( {{4 \over {36}}} \right) + ....\infty$$

$$= {{{4 \over {36}}} \over {1 - {{14} \over {36}}}} = {4 \over {22}} = {2 \over {11}}$$

and P(E) = probability of occurring perfect square before prime

$$= \left( {{7 \over {36}}} \right) + \left( {{{14} \over {36}}} \right)\left( {{7 \over {36}}} \right) + {\left( {{{14} \over {36}}} \right)^2}\left( {{7 \over {36}}} \right) + ....\infty$$

$$= {{{7 \over {36}}} \over {1 - {{14} \over {36}}}} = {7 \over {22}}$$

$$\therefore$$ $$P(T/E) = {{{2 \over {11}}} \over {{7 \over {22}}}} = {4 \over 7} = p \Rightarrow 14p = 8$$

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