calendarBack
Quant

/

Modern Maths

/

Probability
ALL MODULES

CAT 2025 Lesson : Probability - Non-Mutually Exclusive Events

bookmarked

4.2 Non-mutually Exclusive Events

When 22 events are such that they can both occur together, then they are called non-mutually exclusive events.

In other words, two events are not mutually exclusive if there are one or more common outcomes between them, i.e. P(A ∩\cap B) ≠0\ne \bm{0}

Probability of occurrence of either of the two non-mutually exclusive events (A or B) is
P(A or B) == P(A ∪\cup B) = P(A) + P(B) −- P(A ∩\cap B)

Example 7

A number is randomly chosen from the first 100100 natural numbers. What is the probability of the number being a multiple of 33 or 55?

(11) 0.470.47               (22) 0.50.5               (33) 0.530.53               (44) 0.60.6              

Solution

With range being 11 to 100100 (both inclusive),
nn(multiples of 33) =33= 33
nn(multiples of 55) =20= 20

In the above multiples, we have double counted those numbers that are multiples of both 33 and 55. Such numbers will be multiples of LCM(3,5)=15(3, 5) = 15
nn(multiples of 15)=615) = 6

Total number of outcomes =100= 100
Number of favourable outcomes =33+20−6=47= 33 + 20 - 6 = 47

Probability =47100=0.47= \dfrac{47}{100} = 0.47

Answer: (11) 0.470.47

Example 8

What is the probability of a randomly picked card from a pack of 5252 cards being a King or a Spade?

Solution

P(King) =452= \dfrac{4}{52} ; P(Spade) =1352= \dfrac{13}{52} ; P(King & Spade) =152= \dfrac{1}{52}

P(King or Spade) =452+1352−152=1652=413= \dfrac{4}{52} + \dfrac{13}{52} - \dfrac{1}{52} = \dfrac{16}{52} = \dfrac{4}{13}

Answer: 413\dfrac{4}{13}

Example 9

In a group of 2020 students, 88 are girls. 55 girls and 88 boys study biology. If 22 students are randomly selected, what is the probability of they being girls or students who study biology?

(11) 919\dfrac{9}{19}               (22) 1219\dfrac{12}{19}               (33) 4495\dfrac{44}{95}               (44) 4895\dfrac{48}{95}              

Solution

Let A denote the students who are girls and B denote students who study biology.

nn(A) =8= 8, nn(B) =13= 13
nn(A ∩\cap B) =5= 5 [Girls who study biology]
nn(A ∪\cup B) =n= n(A) ++ nn(B) ++ nn(A ∩\cap B) =8+13−5=16= 8 + 13 - 5 = 16

Total number of students =20= 20
P(Both students being girls or study biology) == 16C220C2\dfrac{^{16}C_{2}}{^{20}C_{2}} == 1219\dfrac{12}{19}

Answer: (22) 1219\dfrac{12}{19}

Want to read the full content

Unlock this content & enjoy all the features of the platform

Subscribe Now arrow-right
videovideo-lock