Probability — JEE Main practice

1. Since the mean of the 7 observations is 8:

$$\frac{2+4+10+x+y+12+14}{7}=8 \;\Rightarrow\; x+y=14$$

2. Since the variance is 16, use the formula $\text{Var}=\frac{\sum x_i^2}{n}-(\text{mean})^2$:

$$16=\frac{2^2+4^2+10^2+x^2+y^2+12^2+14^2}{7}-8^2$$

3. Simplifying gives:

$$x^2+y^2=100$$

4. Solving $x+y=14$ and $x^2+y^2=100$ together with $x>y$ gives:

$$x=8,\qquad y=6$$

5. The set $\{1,2,3,x-4,y,5\}$ becomes $\{1,2,3,4,6,5\}=\{1,2,3,4,5,6\}$.

6. We want the probability that the smaller of two chosen numbers is less than 4. It's easier to find the complementary event: both chosen numbers are at least 4 (i.e., from $\{4,5,6\}$).

$$P(\text{both}\ge4)=\frac{{}^3C_2}{{}^6C_2}=\frac{3}{15}$$

7. So the required probability is:

$$1-\frac{3}{15}=\frac{4}{5}$$

Hence, the answer is Option D.

1. Since the probabilities of a distribution must sum to 1, add all the given expressions and set equal to 1:

$$2k+k+3k+2k^2+2k+k^2+k+7k^2=1$$

2. Grouping the linear terms and quadratic terms separately gives:

$$10k^2+9k-1=0$$

3. Factoring:

$$(k+1)(10k-1)=0$$

4. Since $k$ must be a valid (non-negative) probability parameter, we reject $k=-1$ and take:

$$k=\frac{1}{10}$$

5. We want $P(3<x\le6)=P(x=4)+P(x=5)+P(x=6)$:

$$P(3<x\le6)=2k^2+2k+(k^2+k)$$

6. Substituting $k=\frac{1}{10}$:

$$P(3<x\le6)=2\left(\frac{1}{100}\right)+2\left(\frac{1}{10}\right)+\left(\frac{1}{100}+\frac{1}{10}\right)$$

7. Adding these: $P(3<x\le6)=\frac{2}{100}+\frac{20}{100}+\frac{1}{100}+\frac{10}{100}=\frac{33}{100}=0.33$

Hence, the answer is Option A.

1. The total number of ways to choose two distinct numbers from 1 to 50 is:

$$\binom{50}{2}=1225$$

2. The product $ab$ fails to be divisible by 3 only if neither $a$ nor $b$ is divisible by 3. So it's easier to count this complementary (unfavorable) case first.

3. The multiples of 3 from 1 to 50 number:

$$\left\lfloor\frac{50}{3}\right\rfloor=16$$

So the numbers not divisible by 3 number $50-16=34$.

4. The number of unfavorable pairs, where both chosen numbers avoid multiples of 3, is:

$$\binom{34}{2}=\frac{34\times33}{2}=561$$

5. Therefore, the favorable pairs, where the product is divisible by 3, are:

$$1225-561=664$$

6. So the required probability is:

$$\frac{664}{1225}$$

Hence, the answer is Option C.