Probability problems
- 1. 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
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.
- 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
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
- 3. 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
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.
- 4. 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 Discuss
Correct Answer: 1/10
Explanation:
Total 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
- 5. 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
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
- 6. 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
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
- 7. 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
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
- 8. 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
Correct Answer: 4/49
Explanation:
Favourable numbers are 11, 21, 31, 41.
? Required probability = 4/49
- 9. 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
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
- 10. I forgot the last digit of a 7- digit telephone number. If I randomly dial the final 3 digits after correctly dialing the first four, then what is the chance of dialing the correct number?
Options- A. 1/1001
- B. 1/990
- C. 1/999
- D. 1/1000 Discuss
Correct Answer: 1/1000
Explanation:
it is given that last 3 digits are randomly dialed
Then, each of the digit can be selected out of 10 digits in 10 ways. Hence, required probability
= 1/(10)3 = 1/1000
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