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1

### JEE Main 2021 (Online) 27th August Morning Shift

When a certain biased die is rolled, a particular face occurs with probability $${1 \over 6} - x$$ and its opposite face occurs with probability $${1 \over 6} + x$$. All other faces occur with probability $${1 \over 6}$$. Note that opposite faces sum to 7 in any die. If 0 < x < $${1 \over 6}$$, and the probability of obtaining total sum = 7, when such a die is rolled twice, is $${13 \over 96}$$, then the value of x is :
A
$${1 \over 16}$$
B
$${1 \over 8}$$
C
$${1 \over 9}$$
D
$${1 \over 12}$$

## Explanation

Probability of obtaining total sum 7 = probability of getting opposite faces.

Probability of getting opposite faces

$$= 2\left[ {\left( {{1 \over 6} - x} \right)\left( {{1 \over 6} + x} \right) + {1 \over 6} \times {1 \over 6} + {1 \over 6} \times {1 \over 6}} \right]$$

$$\Rightarrow 2\left[ {\left( {{1 \over 6} - x} \right)\left( {{1 \over 6} + x} \right) + {1 \over 6} \times {1 \over 6} + {1 \over 6} \times {1 \over 6}} \right] = {{13} \over {96}}$$ (given)

$$\Rightarrow$$ $$x = {1 \over 8}$$
2

### JEE Main 2021 (Online) 26th August Evening Shift

Two fair dice are thrown. The numbers on them are taken as $$\lambda$$ and $$\mu$$, and a system of linear equations

x + y + z = 5

x + 2y + 3z = $$\mu$$

x + 3y + $$\lambda$$z = 1

is constructed. If p is the probability that the system has a unique solution and q is the probability that the system has no solution, then :
A
$$p = {1 \over 6}$$ and $$q = {1 \over 36}$$
B
$$p = {5 \over 6}$$ and $$q = {5 \over 36}$$
C
$$p = {5 \over 6}$$ and $$q = {1 \over 36}$$
D
$$p = {1 \over 6}$$ and $$q = {5 \over 36}$$

## Explanation

$$D \ne 0 \Rightarrow \left| {\matrix{ 1 & 1 & 1 \cr 1 & 2 & 3 \cr 1 & 3 & \lambda \cr } } \right| \ne 0 \Rightarrow \lambda \ne 5$$

For no solution D = 0 $$\Rightarrow$$ $$\lambda$$ = 5

$${D_1} = \left| {\matrix{ 1 & 1 & 5 \cr 1 & 2 & \mu \cr 1 & 3 & 1 \cr } } \right| \ne 0 \Rightarrow \mu \ne 3$$

$$p = {5 \over 6}$$

$$q = {1 \over 6} \times {5 \over 6} = {5 \over {36}}$$

Option (b).
3

### JEE Main 2021 (Online) 26th August Evening Shift

A fair die is tossed until six is obtained on it. Let x be the number of required tosses, then the conditional probability P(x $$\ge$$ 5 | x > 2) is :
A
$${{125} \over {216}}$$
B
$${{11} \over {36}}$$
C
$${{5} \over {6}}$$
D
$${{25} \over {36}}$$

## Explanation

P(x $$\ge$$ 5 | x > 2) = $${{P(x \ge 5)} \over {P(x > 2)}}$$

= $${{{{\left( {{5 \over 6}} \right)}^4}.{1 \over 6} + {{\left( {{5 \over 6}} \right)}^5}.{1 \over 6} + ....... + \infty } \over {{{\left( {{5 \over 6}} \right)}^2}.{1 \over 6} + {{\left( {{5 \over 6}} \right)}^3}.{1 \over 6} + ...... + \infty }}$$

=$${{{{{{\left( {{5 \over 6}} \right)}^4}.{1 \over 5}} \over {1 - {5 \over 6}}}} \over {{{{{\left( {{5 \over 6}} \right)}^2}.{1 \over 6}} \over {1 - {5 \over 6}}}}} = {\left( {{5 \over 6}} \right)^2} = {{25} \over {36}}$$
4

### JEE Main 2021 (Online) 26th August Morning Shift

Let A and B be independent events such that P(A) = p, P(B) = 2p. The largest value of p, for which P (exactly one of A, B occurs) = $${5 \over 9}$$, is :
A
$${1 \over 3}$$
B
$${2 \over 9}$$
C
$${4 \over 9}$$
D
$${5 \over 12}$$

## Explanation

P (Exactly one of A or B)

$$= P\left( {A \cap \overline B } \right) + \left( {\overline A \cap B} \right) = {5 \over 9}$$

$$= P(A)P(\overline B ) + P(\overline A )P(B) = {5 \over 9}$$

$$\Rightarrow P(A)(1 - P(B)) + (1 - P(A))P(B) = {5 \over 9}$$

$$\Rightarrow p(1 - 2p) + (1 - p)2p = {5 \over 9}$$

$$\Rightarrow 36{p^2} - 27p + 5 = 0$$

$$\Rightarrow p = {1 \over 3}$$ or $${5 \over {12}}$$

$${p_{\max }} = {5 \over {12}}$$

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