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- Question
Ajay and his wife Reshmi appear in an interview for two vaccancies in the same post. The Probability of Ajay's selection is 1/7 and that of his wife Reshmi's selection is 1/5. What is the probability that only one of them will be selected?
Options- A. 5/7
- B. 1/5
- C. 2/7
- D. 2/35
- Correct Answer
- 2/7
ExplanationP( only one of them will be selected) = p[(E and not F) or (F and not E)]
=
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- 1. What is the probability of getting at least one six in a single throw of three unbiased dice?
Options- A. 1/36
- B. 91/256
- C. 13/256
- D. 43/256 Discuss
- 2. What is the probability of an impossible event?
Options- A. 0
- B. -1
- C. 0.1
- D. 1 Discuss
- 3. A five digit number is formed with the digits 0,1,2,3 and 4 without repetition. Find the chance that the number is divisible by 5.
Options- A. 3/5
- B. 1/5
- C. 2/5
- D. 4/5 Discuss
- 4. A coin is tossed 5 times. What is the probability that head appears an odd number of times?
Options- A. 1/2
- B. 1/3
- C. 2/3
- D. 1 Discuss
- 5. When two dice are thrown simultaneously, what is the probability that the sum of the two numbers that turn up is less than 12?
Options- A. 35/36
- B. 17/36
- C. 15/36
- D. 1/36 Discuss
- 6. If a box contains 10 bulbs,of which just three are defective. If a random sample of five bulbs is drawn, find the probability that the sample contains no defective bulbs.
Options- A. 5/12
- B. 7/12
- C. 3/14
- D. 1/12 Discuss
- 7. Out of 17 applicants 8 boys and 9 girls. Two persons are to be selected for the job. Find the probability that at least one of the selected persons will be a girl.
Options- A. 19/34
- B. 5/4
- C. 20/34
- D. 25/34 Discuss
- 8. A bag contains 5 red smileys, 6 yellow smileys and 3 green smileys. If two smileys are picked at random, what is the probability that both are red or both are green in colour?
Options- A. 7
- B. 1/7
- C. 3/7
- D. 0 Discuss
- 9. If the difference between the mode and median is 2, then the difference between the median and mean is:
Options- A. 2
- B. 1
- C. 3
- D. 4 Discuss
- 10. A coin is tossed twice if the coin shows head it is tossed again but if it shows a tail then a die is tossed. If 8 possible outcomes are equally likely. Find the probability that the die shows a number greater than 4, if it is known that the first throw of the coin results in a tail
Options- A. 1/3
- B. 2/3
- C. 2/5
- D. 4/15 Discuss
Probability problems
Search Results
Correct Answer: 91/256
Explanation:
Find the number of cases in which none of the digits show a '6'.
i.e. all three dice show a number other than '6', 5×5×5=125 cases.
Total possible outcomes when three dice are thrown = 216.
The number of outcomes in which at least one die shows a '6' = Total possible outcomes when three dice are thrown - Number of outcomes in which none of them show '6'.
=216?125=91
The required probability = 91/256
Correct Answer: 0
Explanation:
The probability of an impossible event is 0.
The event is known ahead of time to be not possible, therefore by definition in mathematics, the probability is defined to be 0 which means it can never happen.
The probability of a certain event is 1.
Correct Answer: 1/5
Explanation:
Correct Answer: 1/2
Explanation:
The possible outcomes are as follows :
5H, 5T, (H, 4T), (T, 4H), (2H, 3T) (3H, 2T), i.e. 6 outcomes in all.
Therefore the probability that head appears an odd number of times = 3/6 =1/2 (In only three outcomes out of the six outcomes, head appears an odd number of times).
Correct Answer: 35/36
Correct Answer: 1/12
Explanation:
Total number of elementary events =
Number of ways of selecting no defective bulbs i.e., 5 non-defective bulbs out of 7 is .
So,required probability = / = 1/12.
Correct Answer: 25/34
Explanation:
The events of selection of two person is redefined as first is a girl and second is a boy or first is boy and second is a girl or first is a girl and second is a girl.
So the required probability:
=
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Correct Answer: 1/7
Explanation:
Given total number of smileys = 5 + 6 + 3 = 14
Now, required probability =
Correct Answer: 1
Correct Answer: 1/3
Explanation:
Here Sample space S = { HH, HT, T1, T2, T3, T4, T5, T6 }
Let A be the event that the die shows a number greater than 4 and B be the event that the first throw of the coin results in a tail then,
A = { T5, T6 }
B = { T1, T2, T3, T4, T5, T6 }
Therefore, Required probability =
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