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- Question
There are 26 balls marked with alphabetical order A to Z. What is the probability of selecting vowels listed balls?
Options- A. 1
- B. 21/26
- C. 5/26
- D. 5
- Correct Answer
- 5/26
ExplanationWe know that,
Total number of balls n(S) = 26
Number of vowels n(E) = 5
Hence, required probability = n(E)/n(S) = 5/26.
- 1. A bag contains 4 red and 3 black balls. A second bag contains 2 red and 4 black balls. One bag is selected at random. From the selected bag, one ball is drawn. Find the probability that the ball drawn is red.
Options- A. 23/42
- B. 19/42
- C. 7/32
- D. 16/39 Discuss
- 2. A speaks truth in 75% of cases and B in 80% of cases. In what percentage of cases are they likely to contradict each other, narrating the same incident
Options- A. 30/100
- B. 35/100
- C. 45/100
- D. 50/100 Discuss
- 3. A bag contains 50 tickets numbered 1,2,3,4......50 of which five are drawn at random and arranged in ascending order of magnitude.Find the probability that third drawn ticket is equal to 30.
Options- A. 551/15134
- B. 1/2
- C. 552/15379
- D. 1/9 Discuss
- 4. Three unbiased coins are tossed. What is the probability of getting at most two heads?
Options- A. 3/4
- B. 7/8
- C. 1/2
- D. 1/4 Discuss
- 5. A bag contains 2 red Roses, 4 yellow Roses and 6 pink Roses. Two roses are drawn at random. What is the probability that they are not of same color?
Options- A. 1/6
- B. 2/3
- C. 14/33
- D. 5/6 Discuss
- 6. A word consists of 9 letters; 5 consonants and 4 vowels.Three letters are choosen at random. What is the probability that more than one vowel will be selected ?
Options- A. 13/42
- B. 17/42
- C. 5/42
- D. 3/14 Discuss
- 7. Three houses are available in a locality. Three persons apply for the houses. Each applies for one house without consulting others. The probability that all the three apply for the same house is :
Options- A. 2/9
- B. 1/9
- C. 8/9
- D. 7/9 Discuss
- 8. The probabilities that a student will receive an A, B, C or D grade are 0.4, 0.3 , 0.2 and 0.1 respectively. Find the probability that a student will receive Atleast B grade.
Options- A. 0.21
- B. 0.3
- C. 0.7
- D. None of these Discuss
- 9. 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 exactly one defective bulb.
Options- A. 5/12
- B. 7/12
- C. 3/14
- D. 1/12 Discuss
- 10. Two dice are thrown simultaneously. What is the probability of getting two numbers whose product is even?
Options- A. 3/4
- B. 3/8
- C. 5/16
- D. 2/7 Discuss
Probability problems
Search Results
Correct Answer: 19/42
Explanation:
A red ball can be drawn in two mutually exclusive ways
(i) Selecting bag I and then drawing a red ball from it.
(ii) Selecting bag II and then drawing a red ball from it.
Let E1, E2 and A denote the events defined as follows:
E1 = selecting bag I,
E2 = selecting bag II
A = drawing a red ball
Since one of the two bags is selected randomly, therefore
P(E1) = 1/2 and P(E2) = 1/2
Now, = Probability of drawing a red ball when the first bag has been selected = 4/7
= Probability of drawing a red ball when the second bag has been selected = 2/6
Using the law of total probability, we have
P(red ball) = P(A) =
=
Correct Answer: 35/100
Explanation:
Let A = Event that A speaks the truth
B = Event that B speaks the truth
Then P(A) = 75/100 = 3/4
P(B) = 80/100 = 4/5
P(A-lie) = = 1/4
P(B-lie) = = 1/5
Now, A and B contradict each other =[A lies and B true] or [B true and B lies]
= P(A).P(B-lie) + P(A-lie).P(B)
=
= = 35%
Correct Answer: 551/15134
Explanation:
Total number of elementary events =
Given,third ticket =30
=> first and second should come from tickets numbered 1 to 29 = ways and remaining two in ways.
Therfore,favourable number of events =
Hence,required probability = =551 / 15134
Correct Answer: 7/8
Explanation:
Here S = {TTT, TTH, THT, HTT, THH, HTH, HHT, HHH}
Let E = event of getting at most two heads.
Then E = {TTT, TTH, THT, HTT, THH, HTH, HHT}.
P(E) =n(E)/n(S)=7/8.
Correct Answer: 2/3
Explanation:
Possible outcomes = (RY, YP, PR)
2C1 4C1 + 4C1 6C1 + 6C1 2C1
Required probability = (2C1 4C1 + 4C1 6C1 + 6C1 2C1)/(12C2)
= 8 + 24 + 12/66
= 44/66
= 2/3.
Correct Answer: 17/42
Explanation:
3 letters can be choosen out of 9 letters in ways.
More than one vowels ( 2 vowels + 1 consonant or 3 vowels ) can be choosen in ways
Hence,required probability = = 17/42
Correct Answer: 1/9
Explanation:
One person can select one house out of 3= ways =3.
Hence, three persons can select one house out of 3 in 3 x 3 x 3 =9.
Therefore, probability that all thre apply for the same house is 1/9
Correct Answer: 0.7
Explanation:
P(atleast B) = P( B or A) = P(B) + P(A) = (0.3) + (0.4) = 0.7
Correct Answer: 5/12
Explanation:
Total number of elementary events =
Number of ways of selecting exactly one defective bulb out of 3 and 4 non-defective out of 7 is
So,required probability = / = 5/12.
Correct Answer: 3/4
Explanation:
In a simultaneous throw of two dice, we have n(S) = (6 x 6) = 36.
Then, E= {(1, 2), (1, 4), (1, 6), (2, 1), (2, 2), (2, 3), (2, 4), (2, 5), (2, 6), (3, 2), (3,4),(3, 6), (4, 1), (4, 2), (4, 3), (4, 4), (4, 5), (4, 6), (5, 2), (5, 4), (5, 6), (6, 1),6, 2), (6, 3), (6, 4), (6, 5), (6, 6)}
n(E) = 27.
P(E) = n(E)/n(S) = 27/36 = 3/4.
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