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Probability

Probability

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Mutually Exclusive Events
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CAT 2025 Lesson : Probability - Non-Mutually Exclusive Events

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4.2 Non-mutually Exclusive Events

When 222 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}=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 100100100 natural numbers. What is the probability of the number being a multiple of 333 or 555?

(111) 0.470.470.47               (222) 0.50.50.5               (333) 0.530.530.53               (444) 0.60.60.6              

Solution

With range being 111 to 100100100 (both inclusive),
nnn(multiples of 333) =33= 33=33
nnn(multiples of 555) =20= 20=20

In the above multiples, we have double counted those numbers that are multiples of both 333 and 555. Such numbers will be multiples of LCM(3,5)=15(3, 5) = 15(3,5)=15
nnn(multiples of 15)=615) = 615)=6

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

Probability =47100=0.47= \dfrac{47}{100} = 0.47=10047​=0.47

Answer: (111) 0.470.470.47

Example 8

What is the probability of a randomly picked card from a pack of 525252 cards being a King or a Spade?

Solution

P(King) =452= \dfrac{4}{52}=524​ ; P(Spade) =1352= \dfrac{13}{52}=5213​ ; P(King & Spade) =152= \dfrac{1}{52}=521​

P(King or Spade) =452+1352−152=1652=413= \dfrac{4}{52} + \dfrac{13}{52} - \dfrac{1}{52} = \dfrac{16}{52} = \dfrac{4}{13}=524​+5213​−521​=5216​=134​

Answer: 413\dfrac{4}{13}134​

Example 9

In a group of 202020 students, 888 are girls. 555 girls and 888 boys study biology. If 222 students are randomly selected, what is the probability of they being girls or students who study biology?

(111) 919\dfrac{9}{19}199​               (222) 1219\dfrac{12}{19}1912​               (333) 4495\dfrac{44}{95}9544​               (444) 4895\dfrac{48}{95}9548​              

Solution

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

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

Total number of students =20= 20=20
P(Both students being girls or study biology) === 16C220C2\dfrac{^{16}C_{2}}{^{20}C_{2}}20C2​16C2​​ === 1219\dfrac{12}{19}1912​

Answer: (222) 1219\dfrac{12}{19}1912​

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