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SSC CGL Exam  >  SSC CGL Tests  >  Quantitative Aptitude for SSC CGL  >  Test: Probability- 4 - SSC CGL MCQ

Test: Probability- 4 - SSC CGL MCQ


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15 Questions MCQ Test Quantitative Aptitude for SSC CGL - Test: Probability- 4

Test: Probability- 4 for SSC CGL 2024 is part of Quantitative Aptitude for SSC CGL preparation. The Test: Probability- 4 questions and answers have been prepared according to the SSC CGL exam syllabus.The Test: Probability- 4 MCQs are made for SSC CGL 2024 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests for Test: Probability- 4 below.
Solutions of Test: Probability- 4 questions in English are available as part of our Quantitative Aptitude for SSC CGL for SSC CGL & Test: Probability- 4 solutions in Hindi for Quantitative Aptitude for SSC CGL course. Download more important topics, notes, lectures and mock test series for SSC CGL Exam by signing up for free. Attempt Test: Probability- 4 | 15 questions in 15 minutes | Mock test for SSC CGL preparation | Free important questions MCQ to study Quantitative Aptitude for SSC CGL for SSC CGL Exam | Download free PDF with solutions
Test: Probability- 4 - Question 1
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A bag contains four black and five red balls. If three balls from the bag are chosen at random, what is the chance that they are all black?

Detailed Solution for Test: Probability- 4 - Question 1

Black and Black and Black = 4/9 x 3/8 x 2/7 = 23/504 = 1/21.

Test: Probability- 4 - Question 2
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From a pack of 52 playing cards, three cards are drawn at random. Find the probability of drawing a king, a queen and jack.

Detailed Solution for Test: Probability- 4 - Question 2

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Test: Probability- 4 - Question 3
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A group of investigators took a fair sample of 1972 children from the general population and found that there are 1000 boys and 972 girls. If the investigators claim that their research is so accurate that the sex of a new bom child can be predicted based on the ratio of the sample of the population, then what is the expectation in terms of the probability that a new child bom will be a girl?

Detailed Solution for Test: Probability- 4 - Question 3

972/1972 = 243/ 493.

Test: Probability- 4 - Question 4
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A bag contains 6 red, 4 white and 8 blue balls. If three balls are drawn at random, find the probability that (i) all the three balls are of the same colour.
Detailed Solution for Test: Probability- 4 - Question 4

The required probability would be given by: All are Red OR All are white OR All are Blue = (6/18) x (5/17) x (4/16) + (4/18) x (3/17) x (2/16) + (8/18) x (7/17)x (6/16) = 480/(18 x 17 x 16) 5/51 

Test: Probability- 4 - Question 5
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A and B are two candidates seeking admission to the IIMs. The probability that A is selected is 0.5 and the probability that both A and B are selected is at most 0.3. Is it possible that the probability of B getting selected is 0.9.
Detailed Solution for Test: Probability- 4 - Question 5

P (Both are selected) = P(A) x P(B) Since P(A) = 0.5, we get 0.3 = 0.5 x 0.6.
The maximum value of P(B) = 0.6.
Thus P{B) = 0.9 is not possible.

Test: Probability- 4 - Question 6
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The odds in favour of standing first of three students Amit, Vikas and Vivek appearing at an examination are 1 : 2. 2 : 5 and 1 : 7 respectively. What is the probability that either of them will stand first (assume that a tie for the first place is not possible).
Detailed Solution for Test: Probability- 4 - Question 6

P (Amit) = 1/3

P (Vikas) = 2/7

P (Vivek) = 1/8.
Required Probability = 1/3 + 2/7 + 1/8 = 125/168.

Test: Probability- 4 - Question 7
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A and B are two mutually exclusive events of an experiment. If P(A') = 0.65, P(A u B) = 0.65 and P(B) = p , find the value o f p.
Detailed Solution for Test: Probability- 4 - Question 7

P(A) = 1 - 0.65 = 0.35.
Hence, P(B) = 0.65 - 0.35 = 0.3

Test: Probability- 4 - Question 8
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The odds against an event is 5 : 3 and the odds in favour of another independent event is 7 : 5. Find the probability that at least one of the two events will occur.
Detailed Solution for Test: Probability- 4 - Question 8

P (E1) 3/8

P (E2) = 7/12.
Event definition is: E1 occurs and E2 does not occur or E1 occurs and E2 occurs or E2 occurs and E1 does not occur.

(3/8) x (5/12) + (3/8) x (7/12) + (5/8) x (7/12) = 71/96.

Test: Probability- 4 - Question 9
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A husband and a wife appear in an interview for two vacancies for the same post. The probability of husband’s selection is (1/7) and that of the wife’s selection is 1/5. What is the probability that (i) both of them will be selected?
Detailed Solution for Test: Probability- 4 - Question 9

(i) 1/5 x 1/7 = 1/35

Test: Probability- 4 - Question 10
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There are two bags, one of them contains 5 red and 7 white balls and the other 3 red and 12 white balls, and a ball is to be drawn from one or the other of the two bags. Find the chance of drawing a red ball.
Detailed Solution for Test: Probability- 4 - Question 10

The event can be defined as: First bag is selected and red ball is drawn. 1/2 x 5/12 + y2 x 3/15 = (5/24) + (3/30) = 37/120

Test: Probability- 4 - Question 11
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Two fairdices are thrown. Giventh at the sum of the dice is less than or equal to 4, find the probability that only one dice shows two.
Detailed Solution for Test: Probability- 4 - Question 11

The possible outcomes are: (1, 1); (1, 2); (2, 1), (2, 2); (3, 1); (1, 3).
Out of six cases, in two cases there is exactly one ‘2’ Thus, the correct answer is 2/6 =1/3.

Test: Probability- 4 - Question 12
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If 8 coins are tossed, what is the chance that one and only one will turn up Head?
Detailed Solution for Test: Probability- 4 - Question 12

One head and seven tails would have eight positions where the head can come.
Thus, 8 x (1/2)8 = (1/32)

Test: Probability- 4 - Question 13
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Out of all the 2 - digit integers between 1 to 200, a 2-digit number has to be selected at random. What is the probability that the selected number is not divisible by 7?
Detailed Solution for Test: Probability- 4 - Question 13

The count of the event will be given by: The number of all 2 digit integers - the number of all 2 digit integers divisible by 7

Test: Probability- 4 - Question 14
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Tom and Dick are running in the same race; the probability of their winning are 1/5 and 1/2 respectively . Find the probability that (i) either of them will win the race.
Detailed Solution for Test: Probability- 4 - Question 14

(i) 1/5 + 1/2 = 7/10.

Test: Probability- 4 - Question 15
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Amamath appears in an exam that has 4 subjects. The chance he passes an individual subject’s test is 0.8. What is the probability that he will (i) pass in all the subjects?
Detailed Solution for Test: Probability- 4 - Question 15

The event definitions are:

(a) Passes the first AND Passes the second AND Passes the third AND Passes the fourth

(b) Fails the first AND Fails the second AND Fails the third AND Fails the fourth

(c) Fails all is the non-event

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