Q.421. Probability.

Question 421.
A 4" * 4" * 4" cube is painted and then cut into 64 1" * 1" * 1" cubes. A unit cube is then randomly selected and rolled. What is the probability that the top face of the rolled cube is painted. Give answer as a common fraction.

Answer 421.
There are 8 cubes with three faces painted, 24 with two faces pained, 24 with one face painted and 8 with no faces pained.

Let A = event that a cube with three colored faces is selected and rolled
B = event that a cube with two colored faces is selected and rolled
C = event that a cube with one colored faces is selected and rolled and
D = event that a cube with no colored face is selected and rolled
E = event that the randomly selected and rolled cube has top face pained
=> P(A) = 8/64, P(B) = 24/64, P(C) = 24/64 and P(D) = 8/64
and P(E/A) = 1/2, P(E/B) = 1/3, P(E/C) = 1/6 and P(E/D) = 0
=> required probability
= (1/2) * (8/64) + (1/3) * (24/64) + (1/6) * (24/64)
= 4/64 + 8/64 + 4/64
= 1/4.

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Q.316.Probability - Application of Bayes' Theorem

Question 316.
Three urns contain colored balls at follows:
Urn One: 3 red, 4 white, 1 blue
Urn Two: 1 red, 2 white, 3 blue
Urn Three: 4 red, 3 white, 2 blue
One urn is chosen at random and a ball is withdrawn. It turns out to be red. What is the probability that is came from Urn Two?

Answer 316.
Let Ui = the event that urn i is selected
and R = the event that the red ball is selected from the selected urn

As the urns are selected randomly, P(U1) = P(U2) = P(U3) = 1/3

=> P(R/U1) = No. of red balls in urn 1 / total no. of balls in urn 1
                           = 3/8
Similarly, P(R/U2) = 1/6 and P(R/U3) = 4/9

By Bayes' Rule,
P(U2/R)
= P(U2) * P(R/U2) / [P(U1) * P(R/U1) + P(U2) * P(R/U2) + P(U3) * P(R/U3)]
= (1/3) * (1/6) / [(1/3) * (3/8) + (1/3) * (1/6) + (1/3) * (4/9)]
= (1/6) / (3/8 + 1/6 + 4/9)
= (1/6) * (72/71)
= 12/71.

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Q.313. Probability - Use of Calculus to find area

Question 313.
In the game of antidarts a player shoots an arrow into a rectangular board
measuring 6 metres by eight metres.

If the arrow is within one metre of the centre it scores 1 point,
between 1 and 2 metres away it scores 2,
between 2 and 3 metres away it scores 3,
between 3 and 4 metres and yet still on the board it scores 4,
further than 4 metres but still on the board it scores 5.
The arrows always land on the board but otherwise they are purely random.
Show that the probability that he scores exactly 15 points in 3 arrows is given by
(1 -(2/3)arcsin(3/4) - (1/8)sqrt7)^3.

Answer 313.
To score exactly 15 points in 3 arrows, each arrow must score 5 points
=> each arrow must land in the area of the board further than 4 meters from the center,
but still on the board which is the area shaded blue in the figure as given in the following link.

The equation of the circle with radius 4 having center at the center of the board and axes parallel to the rectangular edges of the board is given by
x^2 + y^2 = 16
=> area shaded green
= ∫ x dy ... (y=0 to 3)
= ∫ √(16 - y^2) dy ... (y = 0 to 3)
= [ (y/2) √(16 - y^2) + 8arcsin(y/4)] ... (y = 0 to 3)
= [(3/2) √7 + 8arcsin(3/4)]
=> area shaded blue
= area of the board - 4 * area shaded green
= 48 - 4 * [(3/2) √7 + 8arcsin(3/4)]
= 48 - 6√7 - 32arcsin(3/4)

=> Probability of an arrow scoring 5, P(5)
= area shaded blue / total area of the rectangle
= [48 - 6√7 - 32arcsin(3/4)]/(48)
= 1 - (1/8)√7 - (2/3)arcsin(3/4)
=> Probability of all the 3 arrows landing in blue area and thus scoring exactly a total of 15 points
= [1 - (1/8)√7 - (2/3)arcsin(3/4)]^3.

[Note: The data of arrows scoring 1, 2, 3 and 4 is redundant.]

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Q.312. Probability - Expectation.

Question 312.
Calculate the expected value of the greater of two numbers when two different numbers are picked at random from the numbers 1.......n.

Answer 312.
Number "n" can be selected with lower numbers in (n-1) ways.
Probability of selecting "n", P(n) = (n-1)/nC2
Probability of selecting (n-1) = (n-1)(n-2)/nC2
=> Expectation
= [n(n-1) + (n-1)(n-2) + (n-2)(n-3) +...+ 2*1] / nC2
= [Σ n(n-1) (n=2 to n)] / nC2
= [Σn^2 - Σn ... (n=2 to n)] / nC2
= [(1/6)n(n+1)(2n+1) - 1 - n(n+1)/2 +1]/nC2
= [(n+1)(2n+1) - 3(n+1)] / [3(n-1)]
= (2n^2 - 2) / [3(n-1)]
= 2(n+1)/3.

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Q.260. Probability - selection with and without replacement.

Question 260.
There are 5 red, 3 green and 2 blue balls with numbers 0 or 1 marked on them as under.
R0 R0 R1 R1 R1 G0 G0 G2 B0 B2.
( i ) Two balls are drawn at random with replacement. What is the probability that the sum of the numbers on the two balls is 1?
( ii ) Two balls are drawn at random without replacement. What is the probability that at least one blue ball is drawn?

Answer 260.
( i )
For the sum on the two balls drawn to be 1,
either (i) the first ball has number 0 and the second ball has number 1
or (ii) the first ball has number 1 and the second ball has number 0
=> probability for the sum on two balls drawn with replacement to be 1
= (5/10) * (3/10) + (3/10) * (5/10)
= 3/10.
 ( ii )
For at least one ball to be blue out of two balls drawn simultaneously,
either 1 ball is blue and 1 ball is non-blue
or both balls are blue
=> required probability
= (2C1 * 8C1)/(10C2) + (2C2)/(10C2)
= 17/45.

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Q.231. Conditional Probability.

Question 231.
To investigate the quality of a lot, a sample of 20 lamp is randomly selected. The retailer accepts a lot if there is less than 3 defective lamps. The proportion of defective item is 0.05. Given that the retailer accepts a lot, what is the probability that less than 2 defective lamps were observed?

Answer 231.
Let A = event that the retailer accepts the lot
and B = event that less than 2 defective bulbs are found
This is a conditional probability and
P(B/A) is required.

P(A)
= (0.95)^20 + 20C1 * (0.05) * (0.95)^19 + 20C2 * (0.05)^2 * (0.95)^18
= 0.35849 + 0.37735 + 0.18868
= 0.92452

P(B)
= (0.95)^20 + 20C1 * (0.05) * (0.95)^19
= 0.35849 + 0.37735
= 0.73584

P(B/A)
= P(B∩A) / P(A)
= P(B) / )(A) ... [because B is a subset of (A∩B)]
= (0.73584) / (0.92452)
= 0.7959.

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Q.154. Probability

Question 154.
Let A sub 1,A sub 2, .........A sub n be independent events. Let P(A sub i)= 1/(i+1) for i= 1,2,....n. Find the probability that none of the events occurs.

Answer 154.
For A' compliment of A,
P(A sub i)= 1/(i+1)
=> P(A' sub i) = 1 - 1/(1 +i) = i / (1 +i)
Probability that none of the events occurs
= P[(A' sub 1) ∩ (A' sub 2) ∩ ... ∩ (A' sub n)]
= P(A' sub 1) * P(A' sub 2) * ... * P(A' sub n)
= [1 / (1 + 1)] * [2 / (2 + 1)] * [3 / (3 + 1)] * .... * [n / (n + 1)]
= 1 / (n + 1).

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Q.136. Probability

Question 136.
Suppose that for some experiment, we have three independent events A, B and C.
Suppose that: i) P(AUB) = 0.7 ii) P(C/A) = 0.3 iii) P(B/C) = 0.5.
Find the probability that exactly one of A, B or C occurs.

Answer 136.
As A, B and C are independent events, from ii) and iii),
P(C/A) = P(C) = 0.3 and
P(B/C) = P(B) = 0.5

From i) P(AUB) = 0.7
=> P(A) + P(B) - P(A∩B) = 0.7
=> P(A) + P(B) - P(A)*P(B) = 0.7
......[because A and B are independent events]
=> P(A) + 0.5 - P(A) * (0.5) = 0.7
=> (0.5)P(A) = 0.2
=> P(A) = 0.4
=>
P(A') = 1 - P(A) = 1 - 0.4 = 0.6
P(B') = 1 - P(B) = 1 - 0.5 = 0.5
P(C') = 1 - P(C) = 1 - 0.3 = 0.7

Probability that exactly one of A, B or C occurs is given by
P(A∩B'∩C') + P(B∩A'∩C') + P(C∩A'∩B')
... [because (A∩B'∩C'), (B∩C'∩A') and (C∩A'∩B') are
... mutually exclusive events.]
= P(A)*P(B')*P(C') + P(B)*P(C')*P(A') + P(C)*P(A')*P(B')
... [because A, B and C are independent events]
= (0.4)*(0.5)*(0.7) + (0.5)*(0.7)*(0.6) + (0.3)*(0.6)*(0.5)
= 0.14 + 0.21 + 0.09
= 0.44.

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Q.129. Probability

Question 129.

A speaks the truth 2 out of 3 times and B, 4 out of 5 times. They agree in the assertion that from a bag containing 6 balls of different colours a black ball has been drawn. Find the probability that the statements are true.

Answer 129.

The ball drawn may be black or non-black (any of the remaining 5 different colours) and both may say that it is black.

Let C = event that the ball drawn is black and both agreeing that it is black

P(C) = (1/6)*(2/3)*(4/5) = 4/45.
This is the case in which they are both agreeing and speaking the truth.

Let D = event that the ball drawn is non-black (any of the remaining 5 different colours) and both agreeing that it is black but telling a lie.
Both can tell a lie in 5 different ways. To explain this, suppose the non-black ball drawn is blue, they may say it is black, white, red, yellow, green (all lies) of which probability of telling black is (1/5) on which they agree. A may tell a lie that it is black in (1/5)*(1/3) =1/15 ways and B can tell it in (1/5)*(1/5) =1/25 ways. Also there is (5/6) probability of picking up a non-blackball. Thus, probability that a non-black ball is drawn and both agree that it is black thus asserting that it is black but telling a lie is
P(D) = (5/6)*(1/15)*(1/25) = 1/(450).

Now, we have found the probability of two events in which both are asserting that the ball is black, but when event C occurs the assertion is with speaking the truth and when event D occurs, assertion is by telling a lie.
=> the probability of their asserting that the ball is black and speaking the truth
= P(C) /[P(C) + P(D)]
= (4/45) / [(4/45) + (1/450)]
= 40/41.

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Q.88. Statistics, To prove that rms ≥ arithmetical mean

Question 88.
Prove that the root mean square value ≥ mean value (for any given data).

Answer 88.
For two numbers a and b
(a - b)^2 ≥ 0
=> a^2 + b^2 - 2ab ≥ 0
=> 2a^2 + 2b^2 ≥ a^2 + 2ab + b^2
=> 2 (a^2 + b^2) ≥ ( a + b )^2
=> (a^2 + b^2)/2 ≥[(a + b)/2]^2
=> √[(a^2 + b^2)/2] ≥ (a + b)/2
Thus, for two quantities a and b,
the root mean square value ≥ mean value.

Now, let us see how to prove for 3 quantities.
For 3 quantities, a, b and c,
(a - b)^2 + (b - c)^2 + (c - a)^2 ≥ 0
=> 2(a^2 + b^2 + c^2) ≥ 2 (ab + bc + ca)
=> 3(a^2 + b^2 + c^2) ≥ a^2 + b^2 + c^2 + 2 (ab + bc + ca)
=> 3(a^2 + b^2 + c^2) ≥ (a + b + c)^2
=> (a^2 + b^2 + c^2)/3 ≥ (a + b + c)^2/9
=> (a^2 + b^2 + c^2)/3 ≥ [(a + b + c)/3]^2
=> √[(a^2 + b^2 + c^2)/3] ≥ (a + b + c)/3
Thus, for three quantities a, b and c
the root mean square value ≥ mean value.

From here onwards things start getting more complicated. To explain, let us do for 4 quantities, a, b, c and d.
Now 2 out of these 4 can be selected in 4C2 = 6 ways. Taking all these 6 cominations,
(a - b)^2 + (b - c)^2 + (c - d)^2 + (d - a)^2 + (a -c )^2 + (b - d)^2 ≥ 0
=> 3(a^2 + b^2 + c^2 + d^2) ≥ 2(ab + bc + cd + da + ac + bd)
=> 4(a^2 + b^2 + c^2 + d^2) ≥ a^2 + b^2 + c^2 + d^2 + 2(ab + bc + cd + da + ac + bd)
=> 4(a^2 + b^2 + c^2 + d^2) ≥ (a + b + c + d)^2
=> (a^2 + b^2 + c^2 + d^2)/4 ≥ [(a + b + c + d)/4]^2
=> √[(a^2 + b^2 + c^2 + d^2)/4] ≥ [(a + b + c + d)/4.

Thus, when you have to prove for n quantities, you have to take all possible combinations of 2 out of n quantities, say, a(p) and a(q), then
Σ [ a(p) - a(q) ]^2 ≥ 0, [ p from 1 to n and q from 1 to n, p < q ]
=> (n - 1) Σ [a(p)]^2, (p = 1 to n) ≥ 2 Σ [a(p)a(q)], [ p from 1 to n and q from 1 to n, p < q ]
=> n Σ [a(p)]^2, (p = 1 to n) ≥ Σ [a(p)]^2, (p = 1 to n) + 2 Σ [a(p)a(q)], [ p from 1 to n and q from 1 to n, p < q ]
=> Σ [[a(p)]^2]/n, (p = 1 to n) ≥ [Σ [a(p)]/n]^2, ( p = 1 to n)
Thus, for 'n' quantities,
the root mean square value ≥ mean value

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Q.81. Probability

Question 81.
A batch of experimental electronic components is such that each component has (independently) only a 56% chance of functioning correctly. Two correctly functioning components are required for a space mission, but they can only be tested in orbit. How many of the components must be taken into space to be at least 98% of having two that function?

Answer 81.
Probability of selected component functioning correctly p = 0.56.
Probability of selected component functioning incorrectly q = 0.44.

If n components are selected and at least 2 of them functioning correctly should have a probability ≥ 0.98, then
required probability = 1 - probability that exactly 0 or 1 are correctly functioning
=> 1 - nC0 (0.56)^0 * (0.44)^n - nC1(0.56)^1*(0.44)^(n - 1) ≥ 0.98
=> nC0 (0.56)^0 * (0.44)^n + nC1(0.56)^1*(0.44)^(n - 1) ≤ 0.02
=> (0.44)^n + (0.56n)*(0.44)^(n - 1) ≤ 0.02
=> (0.44)^(n -1) * [ 0.44 + 0.56n ] ≤ 0.02

Using iterative process,
n = 10
=> (0.44)^9 * [0.44 + (0.56)*10] = 0.00373
n = 9
=> (0.44)^8 * [0.44 + (0.56)*9] = 0.0077
n = 8
=> (0.44)^7 * [0.44 + (0.56)*8] = 0.016
n = 7
=> (0.44)^6 * [0.44 + (0.56)*7] = 0.032
Thus, n = 8 is the answer.

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Q.50. Probability, Polynomial function

Question 50.
Suppose f(x)=x^3+ax^+bx+c where a,b,c are chosen respectively by throwing a die three times.?
Then the probability that f(x) is an increasing function, is-

Answer 50.
f(x) = x^3 + ax^2 + bx + c
=> f '(x) = 3x^2 + 2ax + b
For f(x) to be an increasing function, f '(x) ≥ 0
=> 3x^2 + 2ax + b ≥ 0
=> f '(x) has 0 or 1 real root

=> 4a^2 - 12b ≤ 0 => a^2 ≤ 3b
=> b = 1 and a = 1
b = 2 and a = 1, 2
b = 3 and a = 1, 2, 3
b = 4 and a = 1, 2, 3,
b = 5 and a = 1, 2, 3,
b = 6 and a = 1, 2, 3, 4
=> required probab ility
= (1/6) * [(1/6) + (2/6) + 3*(3/6) + (4/6)]
= 4/9.

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Q.32. Probability

AQuestion 32.
From a chess board three blocks are chosen at random. What is the probability that the three blocks are from the same diagonal?

 Answer 32.
Total no. of ways of choosing three blocks out of 64 blocks,
n = 64C3 = 41664.
Three blocks can be chosen from diagonals from left to right in
2 (3C3 + 4C3 + 5C3 + 6C3 + 7C3) + 8C3 ways = 196.
Three blocks can be chosen from diagonals from right to left in 196 ways.
Total number of ways of choosing three blocks such that they are from the same diagonal,
r = 196 + 196 = 392.
By classical definition of probability,
required probability
= r / n
= 392 / 41664
= 7 / 744.

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