One person speaks truth in 60% of the cases and another person in 80% of the cases. They are likely to agree in stating the same fact in Option 1: 48% of the cases Option 2: 56% of the cases Option 3: 58% of the cases Option 4: 70% of the cases

Answer: 56% of the cases. Solution: Person A tells truth in 60% of cases and lies in 40% of cases. Person B tells truth in 80% of cases and lies in 20% of cases. Two people will agree when both tell the truth or both lie. Probability both tell the truth: P(both truth) = 0.60 x 0.80 P(both truth) = 0.48 Probability both lie: P(both lie) = 0.40 x 0.20 P(both lie) = 0.08 Probability they agree: P(agree) = P(both truth) + P(both lie) P(agree) = 0.48 + 0.08 P(agree) = 0.56 Therefore, they are likely to agree in 56% of the cases.

Algebra

Probability

Medium

core

One person speaks truth in 60% of the cases and another person in 80% of the cases. They are likely to agree in stating the same fact in

48% of the cases

56% of the cases

58% of the cases

70% of the cases

✅ Correct Option: 2

Person A tells truth in 60% of cases and lies in 40% of cases.

Person B tells truth in 80% of cases and lies in 20% of cases.

Two people will agree when both tell the truth or both lie.

Probability both tell the truth:

$P(\text{both truth}) = 0.60 \times 0.80$

$P(\text{both truth}) = 0.48$

Probability both lie:

$P(\text{both lie}) = 0.40 \times 0.20$

$P(\text{both lie}) = 0.08$

Probability they agree:

$P(\text{agree}) = P(\text{both truth}) + P(\text{both lie})$

$P(\text{agree}) = 0.48 + 0.08$

$P(\text{agree}) = 0.56$

Therefore, they are likely to agree in 56% of the cases.

2025: 15 May Shift 2

If a random variable $X$ has the following probability distribution:

X	0	1	2	3	4
P(X)	k	2k	3k	k²	6k²

, then

Match List-I with List-II

List-I	List-II
(A) k	(I) 3/7
(B) $P(X < 2)$	(II) 6/49
(C) $P(X > 3)$	(III) 1/7
(D) $P(2 \leq X \leq 3)$	(IV) 22/49

Choose the correct answer from the options given below:

2025: 27 May Shift 1

It is given that 3% of items manufactured by an industry are defective. The probability that a packet of 250 items contains one defective item is: [Given: $e^{-7.5} \approx 0.000553$ ]

2025: 13 May Shift 1

The probability distribution of a ramdom variable $X$ is:

$P(X=x)=\begin{cases}kx^2,& x=1,2,3\\2kx,&x=4,5,6\\0,&\text{otherwise}\end{cases}$ Where $K$ is a constant

Match List-I with List-II

List-I	List-II
(A) $k$	(I) 7/22
(B) $P(X \geq 4)$	(II) 1/44
(C) $P(X < 4)$	(III) 95/22
(D) $E[X]$	(IV) 15/22

Choose the correct answer from the options given below:

2025: 19 May Shift 1

A letter is known to have come either from KOLKATA or TATANAGAR. On the envelope just two consecutive letters TA are visible. The probability that letter has come from TATANAGAR is

2026: 22 May Shift 2

A bag contains 8 red and 7 black balls. Two balls are drawn at random. The probability that both the balls are of the same colour is:

2025: 13 May Shift 2

A random variable X has the following probability distribution

X	0	1	2	3	4	5	6	7	8
P(X)	a	3a	5a	7a	9a	11a	13a	15a	17a

Then the values of 'a' and P(0 < X < 5) respectively are

One person speaks truth in 60% of the cases and another person in 80% of the cases. They are likely to agree in stating the same fact in

Solution

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