Algebra

Probability

Medium

applied

A box contains 10 balls, each marked with one of the digits 0 to 9. If four balls are drawn successively with replacement from the bag, then the probability that none is marked with the digit 0 is:

$\left(\frac{9}{10}\right)^4$

$\left(\frac{1}{10}\right)\left(\frac{9}{10}\right)^3$

$\left(\frac{1}{10}\right)^4$

$\left(\frac{1}{10}\right)^2\left(\frac{9}{10}\right)^2$

✅ Correct Option: 1

The box contains 10 balls marked with digits 0 to 9.

Balls marked with 0: 1 ball

Balls NOT marked with 0: 9 balls (marked 1, 2, 3, 4, 5, 6, 7, 8, 9)

Number of draws: 4 balls, with replacement

With replacement means after each draw, the ball is returned to the box. Each draw has the same conditions:

Total balls available for each draw = 10

For any single draw:

Probability of NOT getting 0 = $\frac{9}{10}$

For none of the 4 draws to show 0, each individual draw must not be 0.

1st draw: Probability = $\frac{9}{10}$

2nd draw: Probability = $\frac{9}{10}$

3rd draw: Probability = $\frac{9}{10}$

4th draw: Probability = $\frac{9}{10}$

Since all events must occur together, multiply the probabilities:

$P(\text{no zero in all 4 draws})$

$= \frac{9}{10} \times \frac{9}{10} \times \frac{9}{10} \times \frac{9}{10}$

$= \left(\frac{9}{10}\right)^4$

Therefore, the probability that none is marked with digit 0 is $\left(\frac{9}{10}\right)^4$

2025: 13 May Shift 2

Which of the following statements are NOT correct about Standard Normal Distribution?

(A) The probability curve of the Standard Normal Distribution is a bell-shaped curve.

(B) The Standard Normal variate (Z) score describes the position of each data point in terms of its distance from the mean, when measured in standard deviation units.

(D) There is a 95.45 % probability of randomly selecting a score between $\mu - \sigma$ and $\mu + \sigma$ , when $\sigma$ is standard deviation and $\mu$ is mean.

Choose the correct answer from the options given below:

2026: 25 May Shift 1

If $P(A) = \dfrac{3}{10}, P(B) = \dfrac{3}{4}$ and $P(A\cap B) = \dfrac{1}{5}$ , then $P(B\mid A)$ is equal to:

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: 15 May Shift 2

If A speaks truth in 75% cases and B speaks truth in 80% cases, then the probability that they contradict each other in a statement, is:

2025: 30 May Shift 1

Match List-I with List-II

Let $A$ and $B$ be two events such that $P(A) = 0.2$ , $P(B) = 0.4$ , $P(B|A) = 0.5$

List-I	List-II
(A) $P(A \cap B)$	(I) $0.5$
(B) $P(A\vert B)$	(II) $0.8$
(C) $P(A \cup B)$	(III) $0.25$
(D) $P(A')$	(IV) $0.1$

Choose the correct answer from the options given below:

2023: 15 June Shift 2

The random variable X has a probability distribution P(X) of the following form, where k is some number.

$P(X=x) = \begin{cases} k, & \text{if } x=0 \\ 2k, & \text{if } x=1 \\ 3k, & \text{if } x=2 \\ 0, & \text{otherwise} \end{cases}$

Then $P(x \leq 2)$ is :

A box contains 10 balls, each marked with one of the digits 0 to 9. If four balls are drawn successively with replacement from the bag, then the probability that none is marked with the digit 0 is:

Solution

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