வெல்லுங்கள் CSAT 2025 -6: Questions & Answers (Mutiple Choice) கொள்குறி வினா- விடைகள் - Probability - நிகழ்தகவு

Questions & Answers (Mutiple Choice)
கொள்குறி வினா- விடைகள்

Probability - நிகழ்தகவு

1. A card is drawn from a well-shuffled pack of cards. What is the probability that is neither a spade nor a queen?
a. 1/26 b. 4/13 c. 17/52 d. 9/13.

2. 2. A speaks truth 75% and B speaks truth 60%. What is the probability that both contradicts each other?
a. 0.5 b. 0.45 c. 0.55 d. 0.65

3. From a bag containing 6 white and 4 red balls. A man draws 4 balls from the bag without replacement. What is the probability of getting atleast one red ball?
a. 1/14 b. 2/5 c. 13/14 d. 1/4.

4. Find the probability of getting head and tail alternatively if a fair coin is tossed 3 times.
a. 1/4 b. 1/3 c. 1/2 d. 1/8.

5. In an examination 70% of students have passed in Statistics and 80% of students have passed in Mathematics. A student is selected at random. What is the probability he passed only Mathematics?
a. 0. b. 0.3 c. 0.2 d. 0.4.

6. The odds in favour of Ramesh to win one game in Chess Tournament is 3 to 4 and the odds in against of Sathya to win the other game in the same Chess Tournament is 5 to 7. They are playing in different games in the tournament. What is the probability that atleast one of them wins the game.
a. 5/21 b. 16/21 c. 1/4 d. 3/4.

7. There are two bags. One bag contains 4 green and 5 yellow balls and the other contains 5 green and 4 yellow balls. One ball is to be drawn from either of the two bags. Find the probability of drawing a yellow ball.
a. 2/9 b. 2/5 c. 1/2 d. 1/9

8. What is the probability that a leap year contains 53 Saturdays?
a. 1/7 b. 3/7 c. 5/7 d. 2/7.

9. In a group of seven boys, two are brothers. Four boys are selected at random, find the probability that the two brothers form a part of four.
a. 2/7 b. 3/7 c. 5/7 d. 4/7.

10. A couple appear in an interview for 2 vacancies. The probability for husband's selection is 1/4 and for wife's selection is 1/3. The probability that only one of them will be selected is
a. 7/12 b. 5/12 c. 1/2 d. 1/12.

11. A can hit a target 3 times in 4 shots, B can hit 3 times in 5 shots whereas C can hit 2 times in 5 shots. When all of them are trying to hit, find the probability that atleast one of them hit the target.
a. 3/50 b. 4/7 c. 47/50 d. 4/21.

12. Two dice are thrown. What is the probability that the sum is a prime number?
a. 2/3 b. 7/12 c. 1/2 d. 5/12

13. What is the probability that the vowels of the word RETAIL always occupy even places of the word?
a. 1/20 b. 1/24 c. 1/30 d. 1/36

14. Shanvi is organising a game where participants select two numbers at random without replacement. From the first six positive integers, let x denotes the larger of the two numbers obtained find the probability that the larger number is 5.
a. 1/3 b. 4/15 c. 1/5 d. 1/6

15. A die is thrown. If E is the event 'the number appearing is even' and 'T' is the event 'the number appearing is a mutiple of three' , then which of the following is true.
a. E and T are mutually exclusive
b. E and T are complement to each other
c. E and T are independent
d. None is correct.

Answers

1. d 2. b 3. c 4. a 5. c

6. b 7. c 8. d 9. a 10. b

11. c 12. d 13. a 14. b 15. c

Explanation to Answers

1. Total number of cards = 52
Number of spade cards = 13
Number of queen cards = 4
Number of spade queen = 1
The probability that the drawn card is either spade or queen is
(13+4 - 1)/52 = 4/13
The probability that the drawn card is neither spade or queen is
1 - 4/13 = 9/13.

2. Let the event that A speaks truth be A
and B speaks truth be B.
So A does not speak truth is A'
and B does not speak truth is B'.
The required event is AB' + A'B.
i.e., A speaks truth while B does not speak truth and vice versa.
Here p(A) = 0.75 ; p(B) = 0.60
Therefore p(A') = 1 - 0.75 = 0.25 and p(B) =1 - 0.60 = 0.40.
But p(AB' + A'B) = p(AB') +p( A'B)
[Since AB' and A'B are mutually exclusive]
= p(A) p(B') + p(A') p(B)
(Here A,B' are independent.And A',B are also independent)
= 0.75×0.40 + 0.25 × 0.60 = 0.45

3. Total number of balls 6 + 4 = 10
The number of ways to select 4 balls out of 10 balls is 10c4.
The contra event of selecting atleast one red ball is no red ball is selected.
ie., All drawn 4 balls are white.
The number of ways to select 4 white balls out of 6 balls is 6c4.
probability for 4 white balls
= (6c4)/(10c4) = 15/210 = 1/14.
Required probability of selecting atleast one red ball is 1 - 1/14 = 13/14.

4. Total number of outcomes is 2^3 = 8
The possible outcomes are
H T H and T H T.
So the required probability is 2/8 = 1/4.

5. % of students passed both Statistics and Mathematics is 70 + 80 - 100 = 50
% of students passed only Mathematics
is 70 - 50 =20
So the required probability is 0.2

6. Probability to win the game by Ramesh is 3/(3+4) = 3/7.
Probability to lose the game by Ramesh is 1 - 3/7 = 4/7.
Probability to lose the other game by Sathya is 5/(5+7) = 5/12.
Probability to win the game by Sathya is 1 - 5/12 = 7/12.
Probability to lose the games by both Ramesh and Sathya is (4/7)×(5/12) = 5/21.
The probability that atleast one of them wins the game is 1 - (5/21) = 16/21.

7. Probability for selecting a bag is 1/2.
Probability for selecting a yellow ball from first bag is 5/9.
Probability for selecting a yellow ball from second bag is 4/9
Probability for selecting a yellow ball from either of the bag is
(1/2)×(5/9) + (1/2)×(4/9)
= (1/2)×(5/9+4/9) = 1/2.

8. Leap year has 366 days.
i.e. 52 weeks and 2 days
That 2 days may be
Sunday + Monday, Monday + Tuesday
Tuesday + Wednesday,
Wednesday + Thursday,
Thursday + Friday, Friday + Saturday and
Saturday + Sunday.
Out 7 possibilities, favour to us is only two.
The probability that a leap year contains 53 Saturdays is 2/7.

9. Out of 7, four will be selected in 7c4 ways. i.e. 7c4 = 7c3 = (7×6×5)/(1×2×3)
= 35.
The brothers must be selected.
So out of five, we have to select only two. The number of ways to select them is 5c2 ways. i.e., (5×4)/(1×2) = 10 ways.
Required probability = 10/35 = 2/7

10. The probabilities of selection of husband and wife respectively are 1/4 and 1/3.
The probabilities of non-selection of husband and wife respectively are
(1 -1/4) and (1 - 1/3). i.e., 3/4 and 2/3.
The probability that husband selected and wife not selected is (1/4)×(2/3) =1/6
The probability that wife selected and husband not selected is (1/3)×(3/4) =1/4
The probability that only one of them will be selected is 1/6 + 1/4 = 5/12.

11. The probabilities of hitting the targets by A, B and C respectively are 3/4, 3/5 and 2/5.
The probabilities of not hitting the targets by A, B and C respectively are 1/4, 2/5 and 3/5.
The probability of not hitting the targets by all A, B and C is
(1/4)×(2/5)×(3/5) = 3/50.
The probability that atleast one of them hit the target is 1 - 3/50 = 47/50.

12. Minimum sum is 2 and Maximum sum is 12.
The prime numbers possible are 2, 3, 5, 7 and 11
The possible outcomes are
(1,1) (1,2) (2,1) (1,4) (2,3) (3,2) (4,1) (1,6)
(2,5) (3,4) (4,3) (5,2) (6,1) (5,6) and (6,5)
Required probability = 15/36 = 5/12

13. In the word RETAIL, there are
Three vowels A, E and I.
And three consonants R, T and L

Odd places can be occupied by the letters R, T and L in 3×2×1 = 6 ways.

Even places can be occupied by the letters A, E and I in 3×2×1 = 6 ways.

Required number of words = 6×6 =36.
Total number of words formed by the six letters = 6! = 720

Hence the probability is = 36/720 = 1/20.

14. The first positive six numbers are
1, 2, 3, 4, 5 and 6.
Total number of selection of two numbers are
(1,2) (1,3) (1,4) (1,5) (1,6)
(2,3) (2,4) (2,5) (2,6)
(3,4) (3,5) (3,6)
(4,5) (4,6)
(5,6) Total selection 15 ways
The possible cases such that the larger of two numbers is 5 are
(1,5) (2,5) (3,5) (4,5)
Required probability = 4/15.

15. The sample space is {1,2,3,4,5,6}

E = {2,4,6} and T = {3,6}
EnT = {6}
P(E) = 3/6 = 1/2 and P(T) = 1/3
P(E nT) = 1/6
Here P(E nT) = P(E).P(T) since
P(E).P(T) = 1/2.1/3 = 1/6 = P(E nT)
Hence E and T are independent.

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