**RD Sharma Solutions Class 11 Maths Chapter 33 Exercise 33.4: **Until now, we’ve computed the likelihood of an event occurring or not occurring using the favourable and total number of elementary events. The **RD Sharma Class 11 Solutions** are described in simple language by topic experts to assist students in improving their analytical thinking and problem-solving abilities. **RD Sharma Solutions Class 11 Maths Chapter 33** Exercise 33.4 pdf can be downloaded for free to help students improve their exam preparation and score higher on their board exams.

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## Download RD Sharma Class 11 Solutions Chapter 33 Probability Exercise 33.4 Free PDF

RD Sharma Solutions Class 11 Maths Chapter 33 Exercise 33.4

### Access RD Sharma Solutions Class 11 Maths Chapter 33 Exercise 33.4- Important Question with Answers

**1. (a)** **If A and B be mutually exclusive events associated with a random experiment such that P (A) = 0.4 and P (B) = 0.5, then find:(i) P(A ∪ B)**

**Solution:**

Given: A and B are two mutually exclusive events.

P (A) = 0.4 and P (B) = 0.5

By definition of mutually exclusive events we know that:

P (A ∪ B) = P (A) + P (B)

Now, we have to find

**(i)** P (A ∪ B) = P (A) + P (B) = 0.5 + 0.4 = 0.9

**(ii)** P (A′ ∩ B′) = P (A ∪ B)′ {using De Morgan’s Law}

P (A′ ∩ B′) = 1 – P (A ∪ B)

= 1 – 0.9

= 0.1

**(iii)** P (A′ ∩ B) [This indicates only the part which is common with B and not A.

Hence this indicates only B]

P (only B) = P (B) – P (A ∩ B)

As A and B are mutually exclusive so they don’t have any common parts.

P (A ∩ B) = 0

∴ P (A′ ∩ B) = P (B) = 0.5

**(iv)** P (A ∩ B′) [This indicates only the part which is common with A and not B.

Hence this indicates only A]

P (only A) = P (A) – P (A ∩ B)

As A and B are mutually exclusive so they don’t have any common parts.

P (A ∩ B) = 0

∴ P (A ∩ B′) = P (A) = 0.4

**(b)** **A and B are two events such that P (A) = 0.54, P (B) = 0.69 and P (A ∩ B) = 0.35. Find (i) ****P (A ∪ B)**

**Solution:**

Given: A and B are two events.

P (A) = 0.54, P (B) = 0.69 and P (A ∩ B) = 0.35

By definition of P (A or B) under the axiomatic approach we know that:

P (A ∪ B) = P (A) + P (B) – P (A ∩ B)

Now we have to find:

**(i)** P (A ∪ B) = P (A) + P (B) – P (A ∩ B)

= 0.54 + 0.69 – 0.35

= 0.88

**(ii)** P (A′ ∩ B′) = P (A ∪ B)′ {using De Morgan’s Law}

P (A′ ∩ B′) = 1 – P (A ∪ B)

= 1 – 0.88

= 0.12

**(iii)** P (A ∩ B′) [This indicates only the part which is common with A and not B.

Hence this indicates only A]

P (only A) = P (A) – P (A ∩ B)

∴ P (A ∩ B′) = P (A) – P (A ∩ B)

= 0.54 – 0.35

= 0.19

**(iv)** P (A′ ∩ B) [This indicates only the part which is common with B and not A.

Hence this indicates only B]

P (only B) = P (B) – P (A ∩ B)

∴ P (A′ ∩ B) = P (B) – P (A ∩ B)

= 0.69 – 0.35

= 0.34

**(c)** **Fill in the blanks in the following table:**

**Solution:**

**(i)** By definition of P (A or B) under axiomatic approach we know that:

P (A ∪ B) = P (A) + P (B) – P (A ∩ B)

By using data from table, we get:

∴ P (A ∪ B) = 1/3 + 1/5 – 1/15

= 8/15 – 1/15

= 7/15

**(ii)** By definition of P (A or B) under axiomatic approach we know that:

P (A ∪ B) = P (A) + P (B) – P (A ∩ B)

P (B) = P (A ∪ B) + P (A ∩ B) – P (A)

By using data from table, we get:

∴ P (B) = 0.6 + 0.25 – 0.35

= 0.5

**(iii)** By definition of P (A or B) under axiomatic approach we know that:

P (A ∪ B) = P (A) + P (B) – P (A ∩ B)

P (A ∩ B) = P (B) + P (A) – P (A ∪ B)

By using data from table, we get:

∴ P (A ∩ B) = 0.5 + 0.35 – 0.7

= 0.15

Hence the table is:

**2. If A and B are two events associated with a random experiment such that P (A) = 0.3, P (B) = 0.4 and P (A ∪ B) = 0.5, find P (A ∩ B).**

**Solution:**

Given: A and B are two events.

P (A) = 0.3, P (B) = 0.5 and P (A ∪ B) = 0.5

Now we need to find P (A ∩ B).

By definition of P (A or B) under axiomatic approach (also called addition theorem) we know that:

P (A ∪ B) = P (A) + P (B) – P (A ∩ B)

So, P (A ∩ B) = P (A) + P (B) – P (A ∪ B)

P (A ∩ B) = 0.3 + 0.4 – 0.5

= 0.7 – 0.5

= 0.2

∴ P (A ∩ B) is 0.2

**3. If A and B are two events associated with a random experiment such that P (A) = 0.5, P (B) = 0.3 and P (A ∩ B) = 0.2, find P (A ∪ B).**

**Solution:**

Given: A and B are two events.

P (A) = 0.5, P (B) = 0.3 and P (A ∩ B) = 0.2

Now we need to find P (A ∪ B).

By definition of P (A or B) under axiomatic approach (also called addition theorem) we know that:

P (A ∪ B) = P (A) + P (B) – P (A ∩ B)

So, P (A ∪ B) = P (A) + P (B) – P (A ∩ B)

P (A ∪ B) = 0.5 + 0.3 – 0.2

= 0.8 – 0.2

= 0.6

∴ P (A ∪ B) is 0.6

**Probability Ex 33.4 Q5****Probability Ex 33.4 Q6****Probability Ex 33.4 Q7****Probability Ex 33.4 Q8****Probability Ex 33.4 Q9****Probability Ex 33.4 Q10****Probability Ex 33.4 Q11****Probability Ex 33.4 Q12****Probability Ex 33.4 Q13****Probability Ex 33.4 Q14****Probability Ex 33.4 Q15****Probability Ex 33.4 Q16****Probability Ex 33.4 Q17****Probability Ex 33.4 Q18****Probability Ex 33.4 Q19****Probability Ex 33.4 Q20****Probability Ex 33.4 Q21****Probability Ex 33.4 Q22****Probability Ex 33.4 Q23****Probability Ex 33.4 Q24****Probability Ex 33.4 Q25****Probability Ex 33.4 Q26****Probability Ex 33.4 Q27****Probability Ex 33.4 Q28**

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## FAQs on RD Sharma Solutions Class 11 Maths Chapter 33 Exercise 33.4

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