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- Question
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
- Correct Answer
- 1/3
ExplanationHere 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 =
- 1. 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
- 2. 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
- 3. 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
- 4. 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
- 5. 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
- 6. 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
- 7. 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
- 8. fifteen persons are sitting around a circular table facing the centre. What is the probability that three particular persons sit together ?
Options- A. 3/91
- B. 2/73
- C. 1/91
- D. 3/73 Discuss
- 9. The first 8 alphabets are written down at random. what is the probability that the letters b,c,d,e always come together ?
Options- A. 1/7
- B. 8!
- C. 7!
- D. 1/14 Discuss
- 10. In a box, there are 9 blue, 6 white and some black stones. A stone is randomly selected and the probability that the stone is black is ¼. Find the total number of stones in the box?
Options- A. 15
- B. 18
- C. 20
- D. 24 Discuss
Probability problems
Search Results
Correct Answer: 1
Correct Answer: 1/7
Explanation:
Given total number of smileys = 5 + 6 + 3 = 14
Now, required probability =
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:
=
=
=
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: 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:
Correct Answer: 3/91
Explanation:
In a circle of n different persons, the total number of arrangements possible = (n - 1)!
Total number of arrangements = n(S) = (15 ? 1)! = 14 !
Taking three persons as a unit, total persons = 13 (in 4 units)
Therefore no. of ways for these 13 persons to around the circular table = (13 - 1)! = 12!
In any given unit, 3 particular person can sit in 3!. Hence total number of ways that any three person can sit =
n(E) = 12! X 3!
Therefore P (E) = probability of three persons sitting together = n(E) / n(S) = 12! X 3 ! / 14!
= 12!x3x2 / 14x13x12! = 6/14x13 = 3/91
Correct Answer: 1/14
Explanation:
The 8 letters can be written in 8! ways.
n(S) = 8!
Let E be the event that the letters b,c,d,e always come together when the first 8 alphabets are written down.
Now the letters a, (bcde), f, g and h can be arranged in 5! ways.
The letters b,c,d and e can be arranged themselves in 4! ways.
n(E) = 5! x 4!
Now, the required P(E) = n(E)/n(S) = 5! x 4!/8! = 1/14
Hence the answer is 1/14.
Correct Answer: 20
Explanation:
We know that, Total probability = 1
Given probability of black stones = 1/4
=> Probability of blue and white stones = 1 - 1/4 = 3/4
But, given blue + white stones = 9 + 6 = 15
Hence,
3/4 ----- 15
1 ----- ?
=> 15 x 4/3 = 20.
Hence, total number of stones in the box = 20.
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