If $M$ and $N$ are any two events, the probability that atleast one of them occurs is
- A$P(M)+P(N)-2 P(M \cap N)$
- ✓$P(M)+P(N)-P(M \cap N)$
- C$P(M)+P(N)+P(M \cap N)$
- D$P(M)+P(N)+2 P(M \cap N)$
Answer: B.
View full solution →29 questions across 7 question groups — pick any mix to generate a Applied Maths paper with step-by-step answer keys.
MCQ
4 Q→022 Marks Questions
8 Q→033 Marks Question
2 Q→045 Marks Questions
7 Q→05Fill in the blanks.
4 Q→064 Marks Questions
3 Q→07Match the following.
1 Q→One sample from each question group in this chapter. Select any group above to see the full set with answer keys.
Answer: B.
View full solution →Answer: C.
View full solution →Answer: C.
View full solution →Answer: B.
View full solution →| Day | Mon | Tue | Wed | Thurs | Fri | Sat |
| Time (a.m.) | 10:35 | 10:20 | 10:22 | 10:27 | 10:25 | 10:40 |
| Column - l | Column - ll |
| (a) If $E_1$ and $E_2$ are the two mutually exclusive events | (i) $E_1 \cap E_2=E_1$ |
| (b) If $E_1$ and $E_2$ are the mutually exclusive and exhaustive events | (ii) $\left(E_1-E_2\right) \cup\left(E_1 \cap E_2\right)=E_1$ |
| (c) If $E_1$ and $E_2$ have common outcomes, then If $E_1$ and $E_2$ are two events such that | (iii) $E_1 \cap E_2=\phi, E_1 \cup E_2=S$, |
| (d) $E_1 \subset E_2$ | (iv) $E_1 \cap E_2=\phi$ |
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