Given that two dice are thrown, the sum of numbers appearing can be any integer from 2 to 12. Let's determine the probability for specific sums.
The pairs that give a sum of 6 are: (1,5), (2,4), (3,3), (4,2), (5,1). Thus, there are 5 successful outcomes. The sample space when throwing two dice is 6×6=36. Therefore, the probability \( P(S=6) \) is \( \frac{5}{36} \).
The pairs that give a sum of 8 are: (2,6), (3,5), (4,4), (5,3), (6,2). Thus, there are also 5 successful outcomes. Therefore, the probability \( P(S=8) \) is also \( \frac{5}{36} \).
The complement probability is used to calculate the probability of the event occurring at least once: If \( A \) is the event that a sum appears at least once in 9 trials, then the complement event \( A' \) is that the sum does not appear in all 9 trials.
Let \( \bar{P_1} \) be the probability that a sum of 6 does not appear in a single trial, so \( \bar{P_1} = 1 - \frac{5}{36} = \frac{31}{36} \). Thus, the probability that 6 does not appear in all 9 trials is \( \left(\frac{31}{36}\right)^9 \). Therefore, the probability that 6 appears at least once, \( P_1 \), is:
\( P_1 = 1 - \left(\frac{31}{36}\right)^9 \).
Similarly, the probability that a sum of 8 does not appear in a single trial is the same, \( \bar{P_2} = \frac{31}{36} \). The probability that 8 does not appear in all 9 trials is \( \left(\frac{31}{36}\right)^9 \). Thus, the probability that 8 appears at least once, \( P_2 \), is:
\( P_2 = 1 - \left(\frac{31}{36}\right)^9 \).
Since \( P_1 \) and \( P_2 \) can be expressed similarly, \( P_1 : P_2 = 1 : 1 \).
| List-I | List-II | ||
|---|---|---|---|
| (A) | $f(x) = \frac{|x+2|}{x+2} , x \ne -2 $ | (I) | $[\frac{1}{3} , 1 ]$ |
| (B) | $(x)=|[x]|,x \in [R$ | (II) | Z |
| (C) | $h(x) = |x - [x]| , x \in [R$ | (III) | W |
| (D) | $f(x) = \frac{1}{2 - \sin 3x} , x \in [R$ | (IV) | [0, 1) |
| (V) | { -1, 1} | ||
| List I | List II | ||
|---|---|---|---|
| (A) | $\lambda=8, \mu \neq 15$ | 1. | Infinitely many solutions |
| (B) | $\lambda \neq 8, \mu \in R$ | 2. | No solution |
| (C) | $\lambda=8, \mu=15$ | 3. | Unique solution |
S is the sample space and A, B are two events of a random experiment. Match the items of List A with the items of List B.
Then the correct match is:
For the probability distribution of a discrete random variable \( X \) as given below, the mean of \( X \) is: