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- Question
The marked price of an article is 40% more than its cost price. If 15% discount is given on the marked price, then what will be the profit percentage?
Options- A. 25
- B. 15
- C. 21
- D. 19
- Correct Answer
- 19
- 1. If four balls are picked at random, what is the probability that two are green and two are blue?
Options- A. 1/18
- B. 1/70
- C. 3/5
- D. 1/2 Discuss
- 2. if from each of the three boxes containing 3 white and 1 black, 2 white and 2 black and 1 white and 3 black balls, one ball is drawn at random, then the probability that 2 white and 1 black ball will be drawn is?
Options- A. 13/32
- B. 1/4
- C. 1/32
- D. 3/16 Discuss
- 3. A box contains 3 white and 2 red balls. If we draw one ball and ball and without replacing the first ball. The probability of drawing red ball in the second drawn is?
Options- A. 8/25
- B. 2/5
- C. 3/5
- D. 21/25 Discuss
- 4. In a ward-robe, Nitish has 3 trousers. One of them is black, second is blue and third brown. In this ward-robe, he has 4 shirts also. One of them is black and the other 3 are white. He opens his ward-robe in the dark and picks out one shirt-trouser pair without examining the colour. What is the likelihood that neither the shirts nor the trousers are black?
Options- A. 1/12
- B. 1/6
- C. 1/4
- D. 1/2 Discuss
- 5. A fair coin is tossed 100 times, The probability of getting head an odd number of times is?
Options- A. 1/4
- B. 2/3
- C. 1/2
- D. 3/4 Discuss
- 6. In a box carrying one dozen of oranges, one third have become bad. If 3 oranges are taken out from the box at random, what is the probability that at least one oranges out of the three oranges picked up is good?
Options- A. 1/55
- B. 54/55
- C. 45/55
- D. 3/55 Discuss
- 7. A fair coin is tossed 100 times . The probability of getting tails an odd number of times is?
Options- A. 1/2
- B. 1/8
- C. 3/8
- D. None of these Discuss
- 8. The probability that 9 students pass in Mathematics, Physics and Chemistry are m, p and c respectively, of these subject the students has a 75% chance of passing in at least one subject 50% chance of passing in at least two subject and 40% chance of passing in exactly two. Which of the following relation are true?
Options- A. p + m + c =19/20
- B. p + m + c = 17/20
- C. pmc = 1/10
- D. pmc = 1/4 Discuss
- 9. If n integer taken at random are multiplied together then the probability that the last digit of the product is 2, 4, 6, 8 is?
Options- A. 2n/5n
- B. 4n - 2n / 5n
- C. 4n/5n
- D. 8n - 4n / 5n Discuss
- 10. A speaks truth in 60% of the cases and B in 80% of the cases. In what percentage of cases are they likely to contradict each other, narrating the same incident?
Options- A. 70%
- B. 36%
- C. 22%
- D. 44% Discuss
More questions
Correct Answer: 1/70
Explanation:
Total number of outcomes = 10c4 = 210
Favourable number of outcomes = 3c2 x 2c2 = 3 x 1 = 3
? Required probability = 3/210 = 1/70
Correct Answer: 13/32
Explanation:
You can have
W, W, B or W, B, W or B, W, W
Reqd. probability
= 3/4. 2/4. 3/4 + 3/4. 2/4. 1/4 + 1/4. 2/4. 1/4
= 9/32 + 3/32 + 1/32 = 13/32
Correct Answer: 2/5
Explanation:
Total balls in the box = 5
Second red ball can be drawn in two ways
Case I. First ball is white and second ball is red.
Its probability = 3/5.2/4 = 6/20 = 3/10
Case II: First ball is red and second ball is red
Its probability = 2/5.1/4 = 2/20 = 1/10
Hence, reqd. probability
= 3/10 + 1/10 = 4/10 = 2/5
Correct Answer: 1/2
Explanation:
Probability that trousers are not black = 2/3
Probability that shirts are not black = 3/4
Required robability = 2/3 x 3/4 = 1/2
Correct Answer: 1/2
Explanation:
n(S) = 2100
n(E) = No. of favourable ways = 100C1 + 100C3 + ... 100 C99 = 2100-1 = 299
[? nC1 + nC3 + nC5 + ........................... = 2n-1 ]
? P(E)= n(E)/n(S) = 299/2100 = 1/2
Note : The given case can be generalised as "If a unbiased coin is tossed 'n' times, then the chance that the head will present itself an odd number of times is 1/2. "
Correct Answer: 54/55
Explanation:
n(S) = 12C3 = (12 x 11 x 10) / (3 x 2) = 2 x 11 x 10 = 220
No. of selecting of 3 oranges out of the total 12 orange = 4C3 = 4
? n(E) = No. of desired selection of oranges = 220 - 4 = 216
? P(E) = n(E) / n(S) = 216 / 220 = 54/55
Correct Answer: 1/2
Explanation:
The total number of cause is 2100.
The number of favourable cases are 100C1 + 100C3 + ..... + 100C99
= 2100 - 1 = 299
? Reqd probability = 299/ 2100 = 1/2
Correct Answer: pmc = 1/10
Explanation:
P(M) = m, P (p) = p, P(c) = c
? The probability of at least one success
= P (M ? P ? C)
= m + p + c -mp - mc - pc + mcp = 3/4 ...(1)
The probability of at least two successes = mcp + mcp + mcp + mcp
= mc(1 - p) + mp (1 - c ) + (1 - m )cp + mcp
= mc + mp + cp - 2mcp = 1/2
The probability of exactly two success
= mcp + mcp + mcp
= mc(1 - p) + mp (1 - c ) cp(1 - m )
= mc + mp + cp - 3 mcp = 2/5
(2) & (3) gives,
? mcp = 1/2 - 2/5 = 1/10
? mc + mp + cp = 2/10 + 1/2 = 1/5 + 1/2 = 7/10
From (1),
m + p + c - 7/10 + 1/10 = 3/4
? m + p + c = 3/4 + 7/10 - 1/10 = 27/20
Thus, pmc = 1/10 is a true relation.
Correct Answer: 4n - 2n / 5n
Explanation:
If the last digit in the product is to 2, 4, 6, 8 the last digit in all the n number should not be 0 and 5 and the last digit of all number should not be selected exclusively from the set of number {1, 3, 7, 9}
? Favourable number of cases
= 8n - 4n
But generally the last digit can be one of 0, 1, 2, 3, .... 9.
Hence, the total number of ways = 10n
Hence, the required probability
= 8n - 4n / 10n
= 4n - 2n / 5n
Correct Answer: 44%
Explanation:
Let A = Event that A speaks the truth
and B = Event that B speaks the truth
Then, P(A) = 60/100 = 3/5 and P(A) = 80/100 = 4/5
? P(A) = (1 - 3/5) = 2/5 and
P(B) = (1 - 4/5) = 1/5
P(A and B contradict each other)
= P [(A speaks the truth and B tells a lie) or (A tells a lie and B speak the truth)]
=P[(A and B) or (A and B)]
=P(A and B) + P (A and B)
= P(A) x P(B) + P(A) x P(B)
= (3/5 x 1/5) + ( 2/5 x 4/5)
= (3/25 + 8/25 ) = 11/25
= (11/25 x 100)% = 44%
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