Aptitude - Probability - Discussion
Discussion Forum : Probability - General Questions (Q.No. 7)
7.
In a class, there are 15 boys and 10 girls. Three students are selected at random. The probability that 1 girl and 2 boys are selected, is:
Answer: Option
Explanation:
Let S be the sample space and E be the event of selecting 1 girl and 2 boys.
| Then, n(S) | = Number ways of selecting 3 students out of 25 | |||
| = 25C3 ` | ||||
|
||||
| = 2300. |
| n(E) | = (10C1 x 15C2) | ||||||
|
|||||||
| = 1050. |
 P(E) = |
n(E) | = | 1050 | = | 21 | . |
| n(S) | 2300 | 46 |
Discussion:
54 comments Page 4 of 6.
Kathir lee said:
9 years ago
@ Sanjana.
Thank you for the explanation.
Thank you for the explanation.
Anony said:
9 years ago
Why is it not solved like this?
Total possible combinations - BBB, GGG, BGB, BBG, GBB, GGB, GBG, BGG i.e. 8.
Fav combination: GBB, BGB, BBG so 3.
So the answer should be 3/8.
Total possible combinations - BBB, GGG, BGB, BBG, GBB, GGB, GBG, BGG i.e. 8.
Fav combination: GBB, BGB, BBG so 3.
So the answer should be 3/8.
Person said:
9 years ago
There are three probabilities, BBG, BGB and GBB. The chance of getting 2 boys and 1 girl is 15/25 X 14/24 X 10/23 which is 7/46. Since there are three ways to get 2 boys and a Girl,
7/46 X 3 = 21/46.
7/46 X 3 = 21/46.
(2)
Archana Mahour said:
9 years ago
Please provide a solution for this question.
A class consists of 80 students, 25 of them are girls and 55 boys, 10 of them are rich and the remaining poor, 20 of them are fair complexion. What is the probability of selecting a fair complexioned rich girl?
A class consists of 80 students, 25 of them are girls and 55 boys, 10 of them are rich and the remaining poor, 20 of them are fair complexion. What is the probability of selecting a fair complexioned rich girl?
Jahid said:
9 years ago
@Nemi,
The equation for solving your problem should be as follows:
Let assume total women in the group is x.
The probability of being a 1st choice woman is x/(x + 5).
2nd choice woman probability is (x - 1)/(x + 5 - 1).
So equation will be,
x/(x + 5) * x - 1/(x + 4) = 3/20.
The equation for solving your problem should be as follows:
Let assume total women in the group is x.
The probability of being a 1st choice woman is x/(x + 5).
2nd choice woman probability is (x - 1)/(x + 5 - 1).
So equation will be,
x/(x + 5) * x - 1/(x + 4) = 3/20.
Jitender said:
9 years ago
@Sanjna you approach is wrong its simple, if they ask to include both the event(Where And used) then use 'x' symbol otherwise if event completing in one of them (Where OR used) then use '+' symbol.
Example: a bag contains 4 white, 5red balls. Find probability :
1) two balls are drawn random of the same colour.
2) two balls are drawn random of different colour.
Ans:
1)Now when two balls is drawn random of the same colour it may be both red 'OR' both white so
total no of balls=9
4C2 + 5C2 / 9C2.
2)Now when two balls is drawn random of different colour then it sure one of them will be red 'AND' one of them will be white so
4C1x5C1 / 9C2.
Total no of balls = 9.
Example: a bag contains 4 white, 5red balls. Find probability :
1) two balls are drawn random of the same colour.
2) two balls are drawn random of different colour.
Ans:
1)Now when two balls is drawn random of the same colour it may be both red 'OR' both white so
total no of balls=9
4C2 + 5C2 / 9C2.
2)Now when two balls is drawn random of different colour then it sure one of them will be red 'AND' one of them will be white so
4C1x5C1 / 9C2.
Total no of balls = 9.
AnnieVal said:
9 years ago
Please solve this: " A school contains 750 boys and 450 girls. If a pupil is chosen at random what is the probability that a girl is chosen?".
Sree said:
9 years ago
It is simple.
Only using the equation (n* (n-1) )/r!. We will get the solution.
Only using the equation (n* (n-1) )/r!. We will get the solution.
Lesly said:
8 years ago
If there are 4 children in a family, what is the probability that the family has an equal number of boys and girls, at least 2 boys?
Please answer it.
Please answer it.
Kuldeep said:
8 years ago
Thanks @Sanjana.
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