BANK & INSURANCE (PROBABILITY) PART 2

Green = G, Black = B and Yellow = Y
It is given that,
p(G) = 1/3 and p(Y) = 3/8

so, we can write,
G/(G + B + Y) = 1/3
G/(G + 7 + Y) = 1/3
3G = G + 7 + Y
2G − Y = 7 …(i)

Similarly,
Y/(G + B + Y) = 3/8
Y/(G + 7 + Y) = 3/8
8Y = 3G + 3Y + 21
5Y − 3G = 21 … (ii)

Solving equation (i) and (ii) we get,
G = 8, B = 7 and Y = 9
Total Number of Yellow balls in the bag = 9

To have an average of 65% , Sandeep needs 260 out of 400.
He has already scored 200 marks in the 1st three quizzes.
That means he has to give atleast 7 correct answers in the last quiz.
So, there can be 3 cases in this.

Case 1: He answers all questions correctly and scores 100 which is possible in only 1 way.

Case 2: He answers 9 correctly with one incorrect.
This is possible in 10 ways as he can give an incorrect answer to any of the 10 questions.

Case 3: He answers 8 questions correctly with 2 incorrect answers. The 2 incorrect answers can be any of the 10 questions which is possible in ¹⁰C₂ ways.

Case 3: He answers 7 questions correctly with 3 incorrect answers. The 3 incorrect answers can be made in ¹⁰C₃ ways.

So, probability = (1 + 10 + ¹⁰C₂ + ¹⁰C₃)/2¹⁰ [we divide by 2¹⁰ because each question has two options and any of the options can be selected in a total of 10 questions]

Required probability = 176/1024 = 11/64

Let the car defective be = D
Let the car non defective = D’

So
P(X/D)
= 200×(0.08) / [200×0.08 + 120×0.03 + 80×0.04]
= 16/16+3.6+3.2
= 0.7

P(Y/D)
= 120×(0.03) / [200×0.08 + 120×0.03 + 80×0.04]
= 3.6/16+3.6+3.2
= 0.16

Ratio = 0.7/0.16

Probability of getting a head or tail = P(H) = P(T)
= 1/2

X looses if he throws a head in 1st or 3rd or 5th or 7th …

Probability of X will lose = 1/2 + (1/2)³ + (1/2)⁵ + …
= 1/2 [1 + (1/2)² + (1/2)⁴ + …]
= 1/2[1/(1 − 1/4)]
= 2/3

Required probability = 1 − 2/3 = 1/3

P(A) = probability that the gift transferred from 1st bag first to second is candy:
= 7/12

P(B) = probability that the gift drawn from second bag is candy

Case I: If chocolate goes to 2nd bag
P(C) = 5/12 × 9/13

Case II: If candy goes to 2nd bag
P(D) = 7/12 × 10/13

P(B) = P(C) + P(D) = (45/156) + (70/156) = 115/156

P(A/B) = Probability of event A when B has
= 7/12 × 10/13 / (115/156) = 14/23

Hence, option c.

Probability of drawing 4 blue and 2 red balls from bag
A = (⁵C₄ × ⁶C₂)/¹¹C₆ = 25/154

Probability of drawing 3 blue and 3 red balls from bag
B = (⁷C₃ × ⁵C₃)/¹²C₆ = 25/66

Required product = 25/154 × 25/66 = 625/10164

⁶C₂/(15 + x)C₂ = 1/14
30 × 14 = (15 + x)(14 + x)
30 × 14 = 210 + 15x + 14x + x²
x² + 29x − 210 = 0
x² + 35x − 6x − 210 = 0
x(x + 35) − 6(x + 35) = 0
(x − 6)(x + 35) = 0
x = 6, −35 (negative value neglected)
∴ x = 6

Possible cases = 3 spade cards + any other card OR 4 spade cards

Required probability
= (¹³C₃ × ³⁹C₁)/⁵²C₄ + (¹³C₄ / ⁵²C₄)

Required probability
= {(286 × 39)/⁵²C₄} + (715/⁵²C₄)

Required probability = (11869/⁵²C₄)

Required probability
= {11869/(52×51×50×49/4×3×2)}

Required probability = {11869/13×17×25×49}

Required probability = {913/17×25×49}

Required probability = (913/20825)

Required probability
= (³C₁ × ³C₁ + ⁶C₁ × ³C₁)/¹²C₂ = 9/22