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- Question
- A card is drawn at random from a pack of 100 cards numbered 1 to 100. The probability drawing a number which is a square is?
Options- A. 1/5
- B. 2/5
- C. 1/10
- D. None of these
- Correct Answer
- 1/10
ExplanationTotal ways = 100
Squares of following no's lie between 1 and 100,
12, 22, 32 , 42, 52 , 62, 72, 82 , 92 , 102
(Which are 10 in numbers.)
So. Required probability = 10/100 = 1/10 - 1. From a set of 17 cards numbered 1, 2, 3 ....... 17 one is drawn at random. The probability that the number is divisible by 3 or 7 is?
Options- A. 2/17
- B. 1/7
- C. 7/17
- D. 10/ 17 Discuss
- 2. When two dice are thrown, the probability that the difference of the number on the dice is 2 or 3 is?
Options- A. 7/18
- B. 3/11
- C. 5/18
- D. 1/2 Discuss
- 3. The probability of occurrence of an event A is 5/9 . The probability of non-occurrence of the event B is 5/11. The probability that at least one of them will occur?
Options- A. 6/11
- B. 5/9
- C. 4/9
- D. 0.8 Discuss
- 4. A six - faced dice is so biased that is twice as likely to show an even number as an odd number when throw. It is thrown twice. The probability that the sum of two number thrown is even is?
Options- A. 1/12
- B. 1/6
- C. 1/3
- D. 5/9 Discuss
- 5. A bag contains 5 blue and 4 black balls. Three balls are drawn at random. What is the probability that 2 are blue and 1 is black?
Options- A. 1/3
- B. 2/5
- C. 1/6
- D. None of these Discuss
- 6. In suffling a pack of card 3 are accidentally dropped then the chance that missing card should be of different suit is?
Options- A. 169/425
- B. 261/425
- C. 104/425
- D. None of these Discuss
- 7. The probability that a card drawn from a pack of 52 cards will from a pack of 52 cards will be diamond or king being to?
Options- A. 4/52
- B. 4/13
- C. 1/52
- D. 2/13 Discuss
- 8. There are four machine and it is know that exactly two of them are faulty, They are tested. One by one in a random order till both the faulty machines are identified. Then the probability that only two tests are needed is?
Options- A. 1/3
- B. 1/6
- C. 1/2
- D. 1/4 Discuss
- 9. If E be the set of all integers with 1 in their unit place. The probability that a number chosen {2, 3, 4,...50} is an elements of E, is?
Options- A. 5/49
- B. 4/49
- C. 3/49
- D. 2/49 Discuss
- 10. Three students are picked at random from a school having a total of 1000 students. The probability that these students will have identical date and month of their birth, is?
Options- A. 3/1000
- B. 3/365
- C. 1/(365)2
- D. None of these Discuss
Probability problems
Search Results
Correct Answer: 7/17
Explanation:
Total number of cases is 17.
? Number divisible by 3 are 3, are 3, 6, 9, 12, 15 (These are 5 in number )
Number divisible by 7 are 7, 14. (These are 2 in number )
There are two favourable number of cases
Total no. of favourable number = 5 + 2
Required probability = 7/17.
Correct Answer: 7/18
Explanation:
The favourable cases are (1, 3), (2, 4) , (3, 5), (4, 6) and (1, 4), (2, 5), (3, 6) and their reversed cases like (3, 1), (4, 2), (5, 3)...
Total number of favourable cases = 2 x 7
? Required Probability P(E) = 14/36 = 7/18
Correct Answer: 0.8
Explanation:
? P(A) = 5/9
? P(A) = 1- 5/9 = 4/9
? P(B) = 5/11
? P(B) = 1 - 5/11 = 6/11
Probability that none of them will occur = P(A ? B) = 4/9 x 5/ 11 = 20/99
Hence, Reqd. probability = 1-20/99 = 79//99
= 0.798 which is near 0.8.
Correct Answer: 5/9
Explanation:
? Probability for odd = p
? Probability for even = 2p
? p + 2p = 1
? 3p = 1
? p = 1/3
? Probability for odd = 1/3, Probability for even = 2/3.
Sum of two nos. is even means either both are odd or both are even
? Reqd. probability = 1/3 x 1/3 + 2/3 x 2/3 = 1/9 + 4/9 = 5/9
[ ? die is thrown twice ]
Correct Answer: 1/6
Explanation:
Let S be the sample space and E be the event of drawing 3 balls out of which 2 are blue and 1 is back .
Then, n (S)= Number of ways of drawing 3 balls out of 9=9c3
= (9 x 8 x 7) / (3 x 2 x 1)
= 84 and
n(E) = Number of ways of drawing 2 balls out of 5 and 1 ball out of 4 .
= 5C2 4C11
= (5 x 4) / (2 x 1) + 4 = 14
? P(E) = n(E)/n(S) = 14/84 = 1/6
Correct Answer: 169/425
Explanation:
Total ways 52C3 = 22100
There are 4 suit in a pack of cards so three suit can be selected in 4C3 ways.
One card each from different unit can selected as = 13C1 X 13C1 X 13C1 ways
So, favourable ways = 4C3 X 13C1 X 13C1 X 13C1 = 8788
? Required probability = 8788/22100 = 169/425
Correct Answer: 4/13
Explanation:
Total ways = 52
There are 13 cards of diamond, 4 cards of king, but one card is king of diamond which is counted both in diamond and king cards
? Favourable ways = 13 + 4 - 1 = 16
? Required probability = 16/52 = 4/13
Correct Answer: 1/6
Explanation:
First we choose 1 machine out of the given 4 , the probability that it is faulty is 2/4 (favorable outcomes/possible outcomes)
Now we have to pick the second faulty machine the probability of doing so is 1/3.
Hence overall probability is 2/4 x 1/3 = 1/6
Correct Answer: 4/49
Explanation:
Favourable numbers are 11, 21, 31, 41.
? Required probability = 4/49
Correct Answer: 1/(365)2
Explanation:
Since, the 1st students can have any day as his birthday and according to the question corresponding to the 1st person and 2nd the 3rd person need to have the same day as their birthday and thus probability that they have identical birthday
= 1x (1/365) x (1/365)
= 1/(365)2
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