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- Question
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
- Correct Answer
- 25/34
ExplanationThe 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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= - 1. 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
- 2. 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 Discuss
- 3. 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
- 4. What is the probability of an impossible event?
Options- A. 0
- B. -1
- C. 0.1
- D. 1 Discuss
- 5. 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
- 6. 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
- 7. 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
- 8. 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
- 9. A room contains 3 brown, 5 black and 4 white chairs. Two chairs are picked and are put in the lawn. What is the probability that none of the chairs picked is white ?
Options- A. 14/33
- B. 14/55
- C. 12/55
- D. 13/33 Discuss
- 10. From a pack of 52 cards, 3 cards are drawn. What is the probability that one is ace, one is queen and one is jack?
Options- A. 19/5525
- B. 16/5525
- C. 17/5525
- D. 7/5525 Discuss
Probability problems
Search Results
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: 2/7
Explanation:
P( only one of them will be selected) = p[(E and not F) or (F and not E)]
=
=
=
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: 1/7
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 =
Correct Answer: 14/33
Explanation:
Total number of chairs = (3 + 5 + 4) = 12.
Let S be the sample space.
Then, n(s)= Number of ways of picking 2 chairs out of 12
= 12×11/2×1 = 66
Let n(E) = number of events of selecting 2 chairs for selecting no white chairs.
=> 8C2 = 8×7/2×1 = 28
Therefore required probability = 28/66 = 14/33.
Correct Answer: 16/5525
Explanation:
Required probability:
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