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Probability

Probability

MODULES

Basics of Probability
Mutually Exclusive Events
Non-Mutually Exclusive Events
Independent Events
Dependent Events
Conditional Probability
Odds For & Against
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Probability 1
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Probability : Level 1
Probability : Level 2
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ALL MODULES

CAT 2025 Lesson : Probability - Independent Events

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5. Event Dependencies

5.1 Independent Events

When 222 or more events occur separately such that the outcome of 111 event does not affect the probability or likelihood of the other, then these events are called independent events.

Probability of occurrence of two independent events (A and B) is
P(A and B) === P(A) ×\times× P(B)

Likewise, probability of both A and B not occurring is
P(A‾\overline{\text {A}}A and B‾)=(1−P(A))×(1−P(B))\overline{\text{B}}) = (1 - \text{P}(\text{A})) \times (1 - \text{P}(\text{B}))B)=(1−P(A))×(1−P(B))

Example 10

The probabilities of Flipdeal, Snapping and Housekart listing their shares in 201520152015 is 0.60.60.6, 0.30.30.3 and 0.40.40.4. What is the probability of at least one of the companies listing their shares in 201520152015?

(111) 0.0720.0720.072            (222) 0.1680.1680.168            (333) 0.8320.8320.832            (444) 0.9280.9280.928           

Solution

P(At least 111 company listing) =1−= 1 -=1− P(No company listing)

P(Flipdeal not listing) =1−0.6=0.4= 1 - 0.6 = 0.4=1−0.6=0.4
P(Snapping not listing) =1−0.3=0.7= 1 - 0.3 = 0.7=1−0.3=0.7
P(Housekart not listing) =1−0.4=0.6= 1 - 0.4 = 0.6=1−0.4=0.6

As each of these companies listing or not are independent events,
P(No company listing) =0.4×0.7×0.6=0.168= 0.4 \times 0.7 \times 0.6 = 0.168=0.4×0.7×0.6=0.168

P(At least 111 company listing) =1−0.168=0.832= 1 - 0.168 = 0.832=1−0.168=0.832

Answer: (333) 0.8320.8320.832

Example 11

A class has 333 students. The probability of Anush, Bhanu and Chandra passing the maths exam is 30%,70%30 \%, 70 \%30%,70% and 80%80 \%80%. What is the probability (in percentage) of
(I) only Anush passing test?
(II) exactly 222 of them passing the test?

Solution

The probability of Anush, Bhanu and Chandra passing the test are 0.3,0.70.3, 0.70.3,0.7 and 0.80.80.8 respectively.

Case I: We need to find the probability of the instance where Anush passes while Bhanu and Chandra fail.

∴\therefore∴ Probability of only Anush passing =0.3×(1−0.7)×(1−0.8)=0.3×0.3×0.2= 0.3 \times (1 - 0.7) \times (1 - 0.8) = 0.3 \times 0.3 \times 0.2=0.3×(1−0.7)×(1−0.8)=0.3×0.3×0.2

=0.018=1.8%= 0.018 = \bm{1.8 \%}=0.018=1.8%

Case II: We need to find the probability of all instances where exactly 222 of them pass and then add them.

Probability of only Anush and Bhanu passing = = =0.3× \times ×0.7× \times ×(1 – 0.8)=0.042 = 0.042 =0.042

Probability of only Bhanu and Chandra passing == =(1 – 0.3)× \times ×0.7× \times ×0.8=0.392 =0.392 =0.392

Probability of only Anush and Chandra passing = = =0.3× \times ×(1 – 0.7)× \times ×0.8=0.072= 0.072 =0.072

Probability of exactly 222 passing =0.042+0.392+0.072=0.506=50.6%= 0.042 + 0.392 + 0.072 = 0.506 = \bm{50.6 \%}=0.042+0.392+0.072=0.506=50.6%

Answer: (I) 1.8%1.8 \%1.8%; (II) 50.6%50.6 \%50.6%

5.1.1 Independent Events with 2 outcomes

If an event has exactly two outcomes −-− A and B −-− such that P(A) +++ P(B) =1= 1=1, and there are nnn such independent events that occur, the probability of getting rrr desired outcomes is

P(A occurring rrr times out of a total nnn ) =(nCr[P(A)]r[P(B)]n−r) = (^{n}\text{C}_{r}\text{[P(A)]}^{r}\text{[P(B)]}^{n - r})=(nCr​[P(A)]r[P(B)]n−r)

Example 12

If 90%90 \%90% of the students in a class pass an exam, what is the probability of exactly 222 out of a randomly selected 444 students to have passed the exam?

Solution

P(Pass) =0.9= 0.9=0.9 and P(Fail) =1−0.9=0.1= 1 - 0.9 = 0.1=1−0.9=0.1

P(222 Pass & 222 Fails) === nCr^{n}C_{r}nCr​[P(Pass)]r^{r}r[P(Fail)]n−r^{n - r}n−r === 4C2(0.9)2(0.1)2^{4}C_{2}(0.9)^{2}(0.1)^{2}4C2​(0.9)2(0.1)2

=6×0.0081=0.0486=4.86%= 6 \times 0.0081 = 0.0486 = \bm{4.86 \%}=6×0.0081=0.0486=4.86%

Answer: 4.86%4.86 \%4.86%

Rationale: If A, B, C and D are the selected students, (0.9)2(0.9)^{2}(0.9)2 (0.1)2(0.1)^{2}(0.1)2 is the probability where a specified two pass while the other two fail. For instance,(0.9)2(0.9)^{2}(0.9)2 (0.1)2(0.1)^{2}(0.1)2 is the probability for A & B passing and C & D failing. However, the two passing students can be selected in 4C2^{4}C_{2}4C2​ ways, which is, therefore, multiplied.

∴\therefore∴ Probability of exactly 2 passing = 4C2(0.9)2(0.1)2^4C_{2} (0.9)^{2} (0.1)^{2}4C2​(0.9)2(0.1)2

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