If P(A B) = 1/2, P(A' B') = 1/3, P(A) = p and P(B) = 2p, then value of p is Option 1: 1/3 Option 2: 1/9 Option 3: 4/9 Option 4: 7/18

Answer: 7/18. Solution: P(A' B')=1-P(A B), so P(A B)=1-13=23. Then P(A)+P(B)=P(A B)+P(A B)=23+12=76. Since P(A)+P(B)=3p, we get p=7/18.

Algebra

Probability

Easy

common

If $P(A \cap B) = \frac{1}{2}$ , $P(A' \cap B') = \frac{1}{3}$ , $P(A) = p$ and $P(B) = 2p$ , then value of $p$ is

$\frac{1}{3}$

$\frac{1}{9}$

$\frac{4}{9}$

$\frac{7}{18}$

✅ Correct Option: 4

$P(A'\cap B')=1-P(A\cup B)$ , so $P(A\cup B)=1-\frac13=\frac23$ . Then $P(A)+P(B)=P(A\cup B)+P(A\cap B)=\frac23+\frac12=\frac76$ . Since $P(A)+P(B)=3p$ , we get $p=\frac{7}{18}$ .

2026: 29 May Shift 2

An insurance company insured 200 scooter drivers, 400 car drivers and 600 Truck drivers. The probability of an accident involving scooter driver, car driver and truck driver are 0.01, 0.03 and 0.15 respectively. One of the insured person meets an accident, then the probability that he is a truck driver is:-

2026: 19 May Shift 1

Two thirds of the students in a class are boys and the rest are girls. It is known that the probability of a girl getting a first class is 0.25 and that of a boy getting a first class is 0.28, then the probability that a student chosen at random will get first class marks in the subject is

2026: 14 May Shift 2

A pair of coins thrown then

Match the LIST-I with LIST-II

	LIST-I		LIST-II
A.	Probability of occurance of head on both coins is	I.	$\dfrac{3}{4}$
B.	Probability of occurance of at least one tail is	II.	$\dfrac{1}{2}$
C.	Probability of occurance of one tail only is	III.	$1$
D.	Probability of a tail appears on one coin given that only one coin shows head.	IV.	$\dfrac{1}{4}$

Choose the correct answer from the options given below:

2026: 22 May Shift 1

A pair of dice is thrown then probability of getting sum of numbers on dice is '8', is:

2022: 17 Aug Shift 2

It is given that only 0.1% of a large population have COVID infection. In this population, the reliability of COVID RTPCR-test is specified as follows :

For persons having COVID, 90% of the test detects the disease but 10% goes undetected. For persons not having COVID, 99% of the test is judged COVID negative but 1% are diagnosed as COVID positive.

Based on the above informations, answer the question :

The probability that the person is actually having COVID given that he is tested as COVID positive is :

2026: 21 May Shift 2

If A and B are two events such that $P(A) = \dfrac{3}{10}, P(B) = \dfrac{2}{5}$ and $P(A \cup B) = \dfrac{3}{5}$ , then $P(B|A) + P(A|B)$ is:

If P(A∩B)=12P(A \cap B) = \frac{1}{2}P(A∩B)=21​, P(A′∩B′)=13P(A' \cap B') = \frac{1}{3}P(A′∩B′)=31​, P(A)=pP(A) = pP(A)=p and P(B)=2pP(B) = 2pP(B)=2p, then value of ppp is

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CUET Mathematics Past Year Questions (Topic-Wise)

If $P(A \cap B) = \frac{1}{2}$ , $P(A' \cap B') = \frac{1}{3}$ , $P(A) = p$ and $P(B) = 2p$ , then value of $p$ is