Three Candidates Were Interviewed by 24 Reviewers 11 Approved of a 8 Approved of B 11
The two events mentioned are independent. The first roll of the die is independent of the 2nd roll. Therefore the probabilities can be directly multiplied.
4) 
A) True
B) False
Solution: (A)
P(AꓵCc) will be just P(A). P(only A)+P(C) will make it P(AꓴC). P(BꓵAcꓵCc) is P(only B) Therefore P(AꓴC) and P(only B) will make P(AꓴBꓴC)
5) Consider a tetrahedral dice and roll it twice. What is the probability that the number on the first roll is strictly college than the number on the 2nd scroll?
Note: A tetrahedral die has only four sides (1, two, 3 and four).
B) three/8
C) 7/sixteen
D) 9/16
Solution: (B)
| (1,1) | (2,1) | (3,1) | (four,1) |
| (1,2) | (two,ii) | (3,2) | (4,ii) |
| (1,3) | (2,3) | (3,3) | (four,3) |
| (ane,4) | (2,four) | (iii,four) | (4,4) |
There are vi out of 16 possibilities where the showtime whorl is strictly higher than the second curlicue.
vi) Which of the post-obit options cannot be the probability of any event?
A) -0.00001
B) 0.5
C) 1.001
A) Only A
B) Just B
C) But C
D) A and B
E) B and C
F) A and C
Solution: (F)
Probability always lie within 0 to 1.
7) Anita randomly picks 4 cards from a deck of 52-cards and places them dorsum into the deck ( Any set of four cards is as likely ). Then, Babita randomly chooses viii cards out of the same deck ( Any ready of 8 cards is equally likely). Presume that the choice of four cards by Anita and the pick of 8 cards by Babita are contained. What is the probability that all 4 cards called past Anita are in the set of viii cards chosen by Babita?
A)48Cfour x 52Civ
B)48C4 x 52C8
C)48Cviii ten 52Ceight
D) None of the to a higher place
Solution: (A)
The total number of possible combination would exist 52 C 4 (For selecting 4 cards by Anita) * 52 C 8 (For selecting 8 cards by Babita).
Since, the four cards that Anita chooses is among the 8 cards which Babita has chosen, thus the number of combinations possible is 52 C 4 (For selecting the 4 cards selected by Anita) * 48 C iv (For selecting whatever other four cards by Babita, since the iv cards selected past Anita are common)
Question Context 8:
A player is randomly dealt a sequence of 13 cards from a deck of 52-cards. All sequences of xiii cards are equally probable. In an equivalent model, the cards are chosen and dealt one at a time. When choosing a card, the dealer is equally likely to pick whatsoever of the cards that remain in the deck.
8) If you dealt xiii cards, what is the probability that the 13th card is a King?
D) 1/12
Solution: (B)
Since we are not told annihilation about the first 12 cards that are dealt, the probability that the 13th card dealt is a King, is the same as the probability that the start card dealt, or in fact any particular card dealt is a Rex, and this equals: 4/52
ix) A fair six-sided die is rolled 6 times. What is the probability of getting all outcomes as unique?
C) 0.23148
D) 0.03333
Solution: (A)
For all the outcomes to exist unique, we accept 6 choices for the starting time turn, five for the 2nd plough, 4 for the third plough and so on
Therefore the probability if getting all unique outcomes will be equal to 0.01543
10) A group of 60 students is randomly split into iii classes of equal size. All partitions are equally likely. Jack and Jill are two students belonging to that group. What is the probability that Jack and Jill volition end upwardly in the same class?
Solution: (B)
Assign a unlike number to each student from 1 to lx. Numbers 1 to twenty go in group i, 21 to forty become to group two, 41 to threescore go to group 3.
All possible partitions are obtained with equal probability past a random assignment if these numbers, it doesn't matter with which students we first, so we are free to showtime by assigning a random number to Jack and so we assign a random number to Jill. Later Jack has been assigned a random number in that location are 59 random numbers bachelor for Jill and 19 of these will put her in the same grouping every bit Jack. Therefore the probability is 19/59
eleven) Nosotros take ii coins, A and B. For each toss of money A, the probability of getting head is ane/two and for each toss of coin B, the probability of getting Heads is 1/3. All tosses of the same money are independent. We select a coin at random and toss it till we go a caput. The probability of selecting coin A is ¼ and coin B is 3/4. What is the expected number of tosses to become the commencement heads?
Solution: (A)
If coin A is selected then the number of times the coin would exist tossed for a guaranteed Heads is 2, similarly, for coin B information technology is iii. Thus the number of times would exist
Tosses = two * (i/4)[probability of selecting coin A] + three*(3/4)[probability of selecting coin B]
= 2.75
12) Suppose a life insurance company sells a $240,000 1 year term life insurance policy to a 25-yr old female for $210. The probability that the female survives the year is .999592. Find the expected value of this policy for the insurance company.
Solution: (C)
P(company loses the money ) = 0.99592
P(company does not lose the money ) = 0.000408
The corporeality of money company loses if it loses = 240,000 – 210 = 239790
While the coin information technology gains is $210
Expected money the company will accept to give = 239790*0.000408 = 97.8
Expect money company gets = 210.
Therefore the value = 210 – 98 = $112
13)
A) True
B) Simulated
Solution: (A)
The above argument is true. Y'all would need to know that
P(A/B) = P(AꓵB)/P(B)
P(C c ꓵA|A) = P(C c ꓵAꓵA)/P(A) = P(C c ꓵA)/P(A)
P(B|A ꓵ C c ) = P(AꓵBꓵC c )/P(A ꓵ C c )
Multiplying the three we would go – P(AꓵBꓵC c ), hence the equations holds true
14) When an event A independent of itself?
A) Always
B) If and only if P(A)=0
C) If and only if P(A)=1
D) If and only if P(A)=0 or 1
Solution: (D)
The result can only be independent of itself when either there is no gamble of it happening or when it is sure to happen. Event A and B is independent when P(AꓵB) = P(A)*P(B). Now if B=A, P(AꓵA) = P(A) when P(A) = 0 or 1.
15) Suppose you lot're in the final circular of "Let'south make a deal" game show and you are supposed to cull from 3 doors – ane, 2 & 3. Ane of the iii doors has a car behind it and other ii doors have goats. Allow'southward say yous choose Door 1 and the host opens Door 3 which has a goat behind it. To assure the probability of your win, which of the post-obit options would you lot choose.
C) It doesn't thing probability of winning or losing is the aforementioned with or without revealing 1 door
Solution: (A)
I would recommend reading this article for a detailed discussion of the Monty Hall's Problem.
sixteen) Cross-fertilizing a blood-red and a white flower produces red flowers 25% of the time. Now we cross-fertilize v pairs of cherry-red and white flowers and produce five offspring. What is the probability that there are no red flower plants in the five offspring?
A) 23.7%
B) 37.2%
C) 22.five%
Solution: (A)
The probability of offspring being Red is 0.25, thus the probability of the offspring non being red is 0.75. Since all the pairs are contained of each other, the probability that all the offsprings are not red would be (0.75) 5 = 0.237. Yous tin can call up of this every bit a binomial with all failures.
17) A roulette wheel has 38 slots – 18 red, 18 black, and ii greenish. You lot play five games and e'er bet on reddish slots. How many games can yous expect to win?
A) 1.1165
B) ii.3684C) ii.6316
C) two.6316
D) four.7368
Solution: (B)
The probability that it would exist Red in any spin is xviii/38. Now, you are playing the game 5 times and all the games are independent of each other. Thus, the number of games that y'all can win would be 5*(eighteen/38) = 2.3684
18) A roulette bicycle has 38 slots, 18 are ruby, xviii are blackness, and 2 are green. You play v games and ever bet on red. What is the probability that you win all the 5 games?
Solution: (B)
The probability that it would be Blood-red in whatsoever spin is 18/38. Now, you are playing for game 5 times and all the games are independent of each other. Thus, the probability that you win all the games is (18/38) 5 = 0.0238
xix) Some exam scores follow a normal distribution with a hateful of xviii and a standard deviation of 6. What proportion of test takers have scored betwixt 18 and 24?
C) 34%
D) None of the to a higher place
Solution: (C)
Then here nosotros would demand to summate the Z scores for value being 18 and 24. Nosotros can hands doing that by putting sample hateful as 18 and population hateful as 18 with σ = six and calculating Z. Similarly nosotros can calculate Z for sample mean as 24.
Z= (X-μ)/σ
Therefore for 26 as X,
Z = (18-18)/vi = 0 , looking at the Z table we find 50% people have scores below 18.
For 24 every bit X
Z = (24-18)/6 = one, looking at the Z tabular array nosotros find 84% people accept scores beneath 24.
Therefore around 34% people take scores between 18 and 24.
20) A jar contains iv marbles. iii Red & one white. Two marbles are drawn with replacement after each draw. What is the probability that the same color marble is drawn twice?
D) 1/8
Solution: (C)
If the marbles are of the same color then it will be three/4 * three/4 + one/iv * 1/4 = 5/8.
21) Which of the following events i southward most probable?
A) At least one 6, when six dice are rolled
B) At least two sixes when 12 die are rolled
C) At to the lowest degree 3 sixes when 18 dice are rolled
D) All the above have aforementioned probability
Solution: (A)
Probability of 'six' turning up in a roll of dice is P(6) = (1/6) & P(6') = (v/half-dozen). Thus, probability of
∞ Case 1: (1/six) * (v/half dozen)5 = 0.06698
∞ Case 2: (one/6)two * (v/6)ten = 0.00448
∞ Case 3: (1/6)iii * (five/half-dozen)xv = 0.0003
Thus, the highest probability is Example one
22) Suppose you were interviewed for a technical role. 50% of the people who sat for the offset interview received the call for second interview. 95% of the people who got a call for 2nd interview felt good most their first interview. 75% of people who did not receive a second call, too felt good about their beginning interview. If y'all felt good later your offset interview, what is the probability that you lot volition receive a 2d interview call?
D) 85%
Solution: (B)
Let'south assume there are 100 people that gave the first circular of interview. The 50 people got the interview phone call for the second round. Out of this 95 % felt adept about their interview, which is 47.5. 50 people did not get a call for the interview; out of which 75% felt practiced about, which is 37.5. Thus, the total number of people that felt good subsequently giving their interview is (37.5 + 47.five) 85. Thus, out of 85 people who felt proficient, merely 47.5 got the call for next round. Hence, the probability of success is (47.5/85) = 0.558.
Another more accustomed way to solve this problem is the Baye's theorem. I leave it to y'all to check for yourself.
23) A coin of bore 1-inches is thrown on a table covered with a grid of lines each ii inches apart. What is the probability that the coin lands within a square without touching any of the lines of the grid? You can assume that the person throwing has no skill in throwing the money and is throwing it randomly.
You can assume that the person throwing has no skill in throwing the coin and is throwing it randomly.
Solution: (B)
Call up about where all the center of the money tin exist when it lands on 2 inches grid and it not touching the lines of the grid.
If the yellow region is a i inch square and the exterior square is of 2 inches. If the eye falls in the yellowish region, the coin volition not impact the grid line. Since the total area is 4 and the surface area of the yellowish region is ane, the probability is ¼ .
24) There area total of 8 bows of 2 each of green, yellow, orange & crimson. In how many means can you select 1 bow?
Solution: (C)
You can select one bow out of 4 different bows, then yous can select one bow in four unlike ways.
25) Consider the following probability density function: What is the probability for X≤six i.east. P(10≤6)
What is the probability for Ten≤six i.e. P(10≤half dozen)
Solution: (B)
To calculate the expanse of a particular region of a probability density office, we need to integrate the function under the bounds of the values for which we need to calculate the probability.
Therefore on integrating the given office from 0 to 6, we get 0.5276
26) In a course of 30 students, approximately what is the probability that two of the students have their altogether on the same day (defined by aforementioned day and calendar month) (assuming it's not a leap twelvemonth)?
For example – Students with birthday 3rd Jan 1993 and tertiary Jan 1994 would be a favorable consequence.
Solution: (C)
The total number of combinations possible for no two persons to have the same birthday in a grade of thirty is thirty * (30-one)/2 = 435.
At present, at that place are 365 days in a year (assuming it's not a bound year). Thus, the probability of people having a different birthday would be 364/365. Now there are 870 combinations possible. Thus, the probability that no two people have the same birthday is (364/365)^ 435 = 0.303.
Thus, the probability that two people would have their birthdays on the same date would be 1 – 0.303 = 0.696
27) Ahmed is playing a lottery game where he must pick two numbers from 0 to ix followed by an English alphabet (from 26-messages). He may choose the same number both times.
If his ticket matches the 2 numbers and i letter drawn in society, he wins the grand prize and receives $10405. If just his letter of the alphabet matches only one or both of the numbers do not match, he wins $100. Under whatever other circumstance, he wins cipher. The game costs him $5 to play. Suppose he has chosen 04R to play.
What is the expected net profit from playing this ticket?
B) $2.81C) $-1.82
C) $-one.82
D) $one.82
Solution: (B)
Expected value in this example
Eastward(X) = P(grand prize)*(10405-5)+P(small)(100-5)+P(losing)*(-5)
P(grand prize)= (ane/ten)*(1/10)*(1/26)
P(small) = 1/26-1/2600, the reason we need to do this is we need to exclude the case where he gets the letter right and besides the numbers rights. Hence, we demand to remove the scenario of getting the letter right.
P(losing ) = ane-ane/26-ane/2600
Therefore nosotros can fit in the values to go the expected value as $ii.81
28) Assume yous sell sandwiches. seventy% people choose egg, and the residue choose craven. What is the probability of selling 2 egg sandwiches to the next 3 customers?
Solution: (C)
The probability of selling Egg sandwich is 0.7 & that of a chicken sandwich is 0.3. Now, the probability that next 3 customers would order ii egg sandwich is 3 * 0.7 * 0.7 *0.three = 0.44. They can order them in whatever sequence, the probabilities would yet be the same .
Question context: 29 – 30
HIV is nonetheless a very scary disease to even get tested for. The US military tests its recruits for HIV when they are recruited. They are tested on three rounds of Elisa( an HIV test) earlier they are termed to be positive.
The prior probability of anyone having HIV is 0.00148. The true positive rate for Elisa is 93% and the true negative rate is 99%.
29) What is the probability that a recruit has HIV, given he tested positive on first Elisa test? The prior probability of anyone having HIV is 0.00148. The truthful positive rate for Elisa is 93% and the truthful negative rate is 99%.
Solution: (A)
I recommend going through the Bayes updating department of this commodity for the understanding of the above question.
30) What is the probability of having HIV, given he tested positive on Elisa the second fourth dimension as well.
The prior probability of anyone having HIV is 0.00148. The true positive rate for Elisa is 93% and the truthful negative rate is 99%.
Solution: (C)
I recommend going through the Bayes updating section of this article for the agreement of the above question.
31) Suppose you're playing a game in which we toss a off-white coin multiple times. You have already lost thrice where you guessed heads merely a tails appeared. Which of the below statements would be correct in this case?
A) You lot should guess heads over again since the tails has already occurred thrice and its more likely for heads to occur now
B) You should say tails because guessing heads is not making you lot win
C) You have the aforementioned probability of winning in guessing either, hence whatever you guess there is just a fifty-50 adventure of winning or losing
D) None of these
Solution: (C)
This is a classic trouble of gambler'due south fallacy/monte carlo fallacy, where the person falsely starts to think that the results should even out in a few turns. The gambler starts to believe that if we accept received 3 heads, you should receive a iii tails. This is however non true. The results would even out only in infinite number of trials.
32) The inference using the frequentist approach will always yield the same effect as the Bayesian approach.
A) Truthful
B) FALSE
Solution: (B)
The frequentist Approach is highly dependent on how we define the hypothesis while Bayesian approach helps u.s. update our prior beliefs. Therefore the frequentist arroyo might result in an reverse inference if nosotros declare the hypothesis differently. Hence the 2 approaches might not yield the same results.
33) Hospital records show that 75% of patients suffering from a disease dice due to that disease. What is the probability that 4 out of the 6 randomly selected patients recover?
Solution: (C)
Call back of this as a binomial since in that location are only ii outcomes, either the patient dies or he survives.
Hither due north =6, and ten=four. p=0.25(probability if living(success)) q = 0.75(probability of dying(failure))
P(X) = n C x p x q north-x = 6 C 4 (0.25) 4 (0.75) 2 = 0.03295
34) The students of a particular form were given 2 tests for evaluation. Twenty-five percent of the class cleared both the tests and forty-five percent of the students were able to clear the first test.
Calculate the percent of students who passed the second test given that they were likewise able to laissez passer the get-go exam.
D) 45%
Solution: (C)
This is a unproblematic problem of conditional probability. Let A exist the event of passing in first test.
B is the issue of passing in the second examination.
P(AꓵB) is passing in both the events
P(passing in 2nd given he passed in the starting time ane) = P(AꓵB)/P(A)
= 0.25/0.45 which is around 55%
35) While it is said that the probabilities of having a boy or a girl are the aforementioned, allow's assume that the bodily probability of having a male child is slightly higher at 0.51. Suppose a couple plans to accept 3 children. What is the probability that exactly ii of them will be boys?
Solution: (A)
Call up of this as a binomial distribution where getting a success is a boy and failure is a girl. Therefore we need to calculate the probability of getting two out of three successes.
P(X) = n C ten p 10 q n-x = iii C 2 (0.51) ii (0.49) 1 = 0.382
36) Heights of 10 year-olds, regardless of gender, closely follow a normal distribution with mean 55 inches and standard deviation 6 inches. Which of the following is true?
A) We would expect more number of 10 year-olds to be shorter than 55 inches than the number of them who are taller than 55 inches
B) Roughly 95% of 10 year-olds are between 37 and 73 inches alpine
C) A ten-year-sometime who is 65 inches tall would be considered more unusual than a 10-year-old who is 45 inches tall
D) None of these
Solution: (D)
None of the in a higher place statements are true.
37) About xxx% of human being twins are identical, and the rest are fraternal. Identical twins are necessarily the aforementioned sex, one-half are males and the other half are females. I-quarter of fraternal twins are both males, one-quarter both female, and one-half are mixed: one male person, one female. Y'all have just go a parent of twins and are told they are both girls. Given this information, what is the probability that they are identical?
A) 50%
B) 72%
C) 46%
D) 33%
Solution: (C)
This is a classic problem of Bayes theorem.
P(I) denoted Probability of existence identical and P(~I) denotes Probability of not being identical
P(Identical) = 0.three
P(not Identical)= 0.7
P(FF|I)= 0.5
P(MM|I)= 0.5
P(MM|~I)= 0.25
P(FF|~I)= 0.25
P(FM|~I)= 0.25
P(I|FF) = 0.46
38) Rob has fever and the md suspects information technology to exist typhoid. To be sure, the medico wants to conduct the examination. The test results positive when the patient actually has typhoid 80% of the time. The test gives positive when the patient does non take typhoid ten% of the time. If 1% of the population has typhoid, what is the probability that Rob has typhoid provided he tested positive?
Solution: (B)
We demand to notice the probability of having typhoid given he tested positive.
=P(testing +ve and having typhoid) / P(testing positive)
= = 0.074
39) Jack is having two coins in his hand. Out of the two coins, one is a real coin and the second one is a faulty one with Tails on both sides. He blindfolds himself to choose a random coin and tosses it in the air. The coin falls downwards with Tails facing upward. What is the probability that this tail is shown past the faulty money?
D) 1/4
Solution: (B)
We need to observe the probability of the coin beingness faulty given that information technology showed tails.
P(Faulty) = 0.five
P(getting tails) = 3/four
P(faulty and tails) =0.5*1 = 0.v
Therefore the probability of money existence faulty given that it showed tails would be ii/3
40) A fly has a life between 4-6 days. What is the probability that the wing will die at exactly 5 days?
A) one/2
B) 1/4
C) 1/3
D) 0
Solution: (D)
Here since the probabilities are continuous, the probabilities course a mass role. The probability of a certain outcome is calculated past finding the surface area under the curve for the given weather condition. Here since nosotros're trying to calculate the probability of the fly dying at exactly 5 days – the area under the curve would exist 0. As well to come up to think of it, the probability if dying at exactly 5 days is incommunicable for us to even figure out since nosotros cannot mensurate with space precision if it was exactly 5 days.
Stop Notes
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Source: https://www.analyticsvidhya.com/blog/2017/04/40-questions-on-probability-for-all-aspiring-data-scientists/
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