Aptitude - Probability - Discussion
Discussion Forum : Probability - General Questions (Q.No. 8)
8.
In a lottery, there are 10 prizes and 25 blanks. A lottery is drawn at random. What is the probability of getting a prize?
Answer: Option
Explanation:
| P (getting a prize) = | 10 | = | 10 | = | 2 | . |
| (10 + 25) | 35 | 7 |
Discussion:
25 comments Page 2 of 3.
Priya said:
9 years ago
Total will be 35 as 10 prizes + 25 blanks.
Megala said:
1 decade ago
What is the meaning of blanks and how could be the total sample space is 35?
Saheeba batra said:
1 decade ago
Hi Amulya is the answer 1/10?
If yes then, first draw a hexagon ABCDEF, you will see that te only equilateral triangles than can be formed are ACE and FBD, order of the points wont matter as it leads to the same triangle.
Now the number of ways in which points can be chosen are 6c3.
So your answer is : 2/6c3.
If yes then, first draw a hexagon ABCDEF, you will see that te only equilateral triangles than can be formed are ACE and FBD, order of the points wont matter as it leads to the same triangle.
Now the number of ways in which points can be chosen are 6c3.
So your answer is : 2/6c3.
Aditi said:
1 decade ago
What is the meaning of blank in this question?
Amulya said:
1 decade ago
Please I need help !
Three of six vertices of a rectangle hexagon are chosen at random. The possibility that the triangle with three vertices is equilateral equal to ?
Three of six vertices of a rectangle hexagon are chosen at random. The possibility that the triangle with three vertices is equilateral equal to ?
UmaShankar said:
1 decade ago
n(s)=10c29(choosing two parts)
n(e)=10c1*9c1 (Becoz we are not replacing the part after selecting so first from 10 parts 1 is selected and from remaining 9 one is selected as 9c1)
n(e)=10c1*9c1 (Becoz we are not replacing the part after selecting so first from 10 parts 1 is selected and from remaining 9 one is selected as 9c1)
Victor said:
1 decade ago
Please I need help with this problem!
In a box of 10 electrical parts, 2 are defective. If two parts are chosen at random from the box without replacement, what is the probability that both are defective?
In a box of 10 electrical parts, 2 are defective. If two parts are chosen at random from the box without replacement, what is the probability that both are defective?
Santosh said:
1 decade ago
Here totally we have 10p and 25b (like red and blue balls). So totally we have 35. We need only prizes then 10c1/35c1.
Vinay Rajagopal said:
1 decade ago
@Shree Hemalatha & @Abhijeet :
It is a very cleverly worded problem. The 1st sentence "In a lottery, there are 10 prizes and 25 blanks" implies that there are a total of 10 prizes plus (+) the 25 blanks so Total of 35 :-D
Therefore, solution would be 10 / 35 = 2 / 7!
It is a very cleverly worded problem. The 1st sentence "In a lottery, there are 10 prizes and 25 blanks" implies that there are a total of 10 prizes plus (+) the 25 blanks so Total of 35 :-D
Therefore, solution would be 10 / 35 = 2 / 7!
Garima gupta said:
1 decade ago
n(s)total nbr cases:35c1 = 10+25=35
event happen n(e):10c1=10
(for geting one price frm the nbr of possibilities)
probability =n(e)/n(s)
=10/35
=2/7
event happen n(e):10c1=10
(for geting one price frm the nbr of possibilities)
probability =n(e)/n(s)
=10/35
=2/7
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