Algebra

Probability

Medium

applied

There are $6$ cards numbered $1$ to $6$ , one number on one card. Two cards are drawn at random without replacement. Let X denote the sum of the numbers on the two cards drawn. Then $\mathrm{P}(\mathrm{X}>3)$ is :

$\frac{14}{15}$

$\frac{1}{15}$

$\frac{11}{12}$

$\frac{1}{12}$

✅ Correct Option: 1

Step-by-Step Solution:

Step 1: Find the total number of outcomes

Two cards are drawn at random without replacement from 6 cards. The total number of ways to choose 2 cards out of 6 is given by the combination formula:

$\text{Total Outcomes} = \binom{6}{2} = \frac{6 \times 5}{2 \times 1} = 15$

Step 2: Find the favorable outcomes for $P(X \le 3)$

Instead of counting all pairs where the sum is greater than 3, it is much quicker to use the complement method:

$P(X > 3) = 1 - P(X \le 3)$

Let's find the pairs that give a sum of 3 or less ( $X \le 3$ ):

The smallest possible sum from two distinct cards is $1 + 2 = 3$ .
There are no other combinations that give a sum equal to or less than 3.

Thus, there is only 1 favorable pair: $\{(1, 2)\}$ .

Step 3: Calculate the probability

The probability of getting a sum less than or equal to 3 is:

$P(X \le 3) = \frac{1}{15}$

Now, subtract this from 1 to find $P(X > 3)$ :

$P(X > 3) = 1 - \frac{1}{15} = \frac{14}{15}$

2026: 19 May Shift 2

A beg contains 4 white and 6 red balls. Four balls are drawn at random, then match the probability of the numbers of white balls drawn. Where P(E) denotes the probability of an event E.

Then Match List-I with List-II

LIST-I		LIST-II
A.	P (no white ball)	I.	$\frac{8}{21}$
B.	P (all white balls)	II.	$\frac{1}{14}$
C.	P (one white ball)	III.	$\frac{6}{14}$
D.	P (two white balls)	IV.	$\frac{1}{210}$

Choose the correct answer from the options given below:

2022: 7 Aug Shift 2

A shopkeeper sells three types of flower seeds $A_1, A_2, A_3$ . They are sold as a mixture where the proportions are 4 : 4 : 2 respectively. The germination rates of the three types of seeds are 45%, 60% and 35% respectively. Calculate the probability in the following cases.

The probability that seed will not germinate, given that the seed is of type $A_3$ .

2026: 22 May Shift 2

Five bad apples are mixed with 15 good apples accidently. Let the random variable X be the number of bad apples in a draw of two apples, then mean of X is

2025: 30 May Shift 1

The probability that A hits a target is $\frac{1}{5}$ and the probability that B hits it is $\frac{2}{3}$ . The probability that the target will be hit if both A and B shoot at it independently is:

2026: 29 May Shift 1

If E and F are two events such that $P(E \cup F)=P(E)$ , then

2025: 26 May Shift 2

In a Binomial distribution, the probability of getting a success is $\frac{3}{4}$ and the variance is $\frac{3}{8}$ then the probability of no success is:

There are 666 cards numbered 111 to 666, one number on one card. Two cards are drawn at random without replacement. Let X denote the sum of the numbers on the two cards drawn. Then P(X>3)\mathrm{P}(\mathrm{X}>3)P(X>3) is :

Solution

Step-by-Step Solution:

Step 1: Find the total number of outcomes

Step 2: Find the favorable outcomes for P(X≤3)P(X \le 3)P(X≤3)

Step 3: Calculate the probability

Related questions:

Keyboard Shortcuts

CUET Mathematics Past Year Questions (Topic-Wise)

There are $6$ cards numbered $1$ to $6$ , one number on one card. Two cards are drawn at random without replacement. Let X denote the sum of the numbers on the two cards drawn. Then $\mathrm{P}(\mathrm{X}>3)$ is :

Step 2: Find the favorable outcomes for $P(X \le 3)$