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 1 of 3.
Shree Hemalatha said:
1 decade ago
I cant understand the concept, could you explain me clearly.
Arumugam said:
1 decade ago
n(s)=35c1
n(e)=10c1
ans=10/35=2/7.
n(e)=10c1
ans=10/35=2/7.
Thameem ansari said:
1 decade ago
Total samples n(s)=25+10=35;
for getting one prize from 10 prizes n(e)=10c1;
probability =n(s)/n(e)=10/35=2/7
for getting one prize from 10 prizes n(e)=10c1;
probability =n(s)/n(e)=10/35=2/7
Abhijeet said:
1 decade ago
I can't understand concept.
Swetha said:
1 decade ago
Probability= n(e)/n(s)
n(e)= 10c1 =10
n(s)= 10+25 =35c1 =35
p=10/35=2/7
n(e)= 10c1 =10
n(s)= 10+25 =35c1 =35
p=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
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!
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.
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?
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)
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