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 1 of 6.
Daniel ocen said:
3 months ago
I can't understand this, someone help me to get it.
Varshini said:
11 months ago
Total (boys and girls)==>25.
Girls can be selected in 10 ways ==>10/25.
Boys can be selected in 15 ways ==>15/24 * 14/23.
Here boys are repeated twice so we can write the total equation like this;
==> (10/25)*(15/24)*(14/23)*(3!/2!).
So by solving the above equation, we get ==> 21/46.
Note: Why I am taking 3!/2! means total 3 students are selected so ==>3! and boys repeated twice so divide by ==> 2!.
Girls can be selected in 10 ways ==>10/25.
Boys can be selected in 15 ways ==>15/24 * 14/23.
Here boys are repeated twice so we can write the total equation like this;
==> (10/25)*(15/24)*(14/23)*(3!/2!).
So by solving the above equation, we get ==> 21/46.
Note: Why I am taking 3!/2! means total 3 students are selected so ==>3! and boys repeated twice so divide by ==> 2!.
(7)
DINESH D S said:
2 years ago
@All.
Here is the simple explanation.
There are 3 possiblities (gbb, bbg, bgb)
Let's take 1st one GBB
Getting girl as prob is 10/25.
Getting boy as prob is 15/24 (as 1 is already taken as girl..sample space is 24).
Getting 2nd boy is 14/23(as 1 boy is already chosen (numerator and denominator(samplw space) are reduced by 1))
Since 3 possibilities 3*(10/25*15/24*14/23).
Will give you the answer 21/46.
Here is the simple explanation.
There are 3 possiblities (gbb, bbg, bgb)
Let's take 1st one GBB
Getting girl as prob is 10/25.
Getting boy as prob is 15/24 (as 1 is already taken as girl..sample space is 24).
Getting 2nd boy is 14/23(as 1 boy is already chosen (numerator and denominator(samplw space) are reduced by 1))
Since 3 possibilities 3*(10/25*15/24*14/23).
Will give you the answer 21/46.
(37)
Ronald said:
2 years ago
I don't understand the answer, please someone help me.
(7)
Karthick M said:
3 years ago
@Prabhu and @Priyaa.
The question is given as 1 girl and 2 boys. Given 'AND' we use simple of multiplication.
Suppose they are given In the question if 1 girl or 2 girls. We use simple subtraction.
That's all.
The question is given as 1 girl and 2 boys. Given 'AND' we use simple of multiplication.
Suppose they are given In the question if 1 girl or 2 girls. We use simple subtraction.
That's all.
(8)
Thanos said:
5 years ago
@Jay Lee Lord.
Thank You for your answer.
Thank You for your answer.
K Pradeep Kumar Reddy said:
6 years ago
What is the mistake in my approach?
Probability of selecting a girl = 10/25.
Probability of selecting a boy = 15/25.
Probability of selecting one girl and two boys is 10/25 + 2* 15/25 = 40/25 = 8/5.
Totally there are three combinations of selecting 1 Girl and 2 Boys.
GBB
BGB
BBG
So the total probability of selecting one girl and two boys is 3*8/5 = 24/5.
Probability of selecting a girl = 10/25.
Probability of selecting a boy = 15/25.
Probability of selecting one girl and two boys is 10/25 + 2* 15/25 = 40/25 = 8/5.
Totally there are three combinations of selecting 1 Girl and 2 Boys.
GBB
BGB
BBG
So the total probability of selecting one girl and two boys is 3*8/5 = 24/5.
(7)
Jeevan said:
7 years ago
@Mahendra.
See, we can select 1girl and 3 boys in 3 ways{gbb, bgb, bbg} and you have followed only one way which is {gbb}. It is not a correct way.
Here, we have to Find the probability for 3 ways then sum those results, you will get 21/46 or As for every way it gives the result as 7/46 you can multiply it with 3 then you will get 21/46.
See, we can select 1girl and 3 boys in 3 ways{gbb, bgb, bbg} and you have followed only one way which is {gbb}. It is not a correct way.
Here, we have to Find the probability for 3 ways then sum those results, you will get 21/46 or As for every way it gives the result as 7/46 you can multiply it with 3 then you will get 21/46.
(2)
Mahendra said:
7 years ago
But without applying the formula, trying to solve it, gives me a different answer.
The probability of selecting girl = 10/25.
Next, Probability of selecting 1 boy = 15/24 (24 because 1 girl is already selected).
Next, the Probability of selecting another boy = 14/23 (23 because 1 boy and 1 girl were already selected) (14 because 1 boy was already selected before).
So,
P(E) = n(E)/n(s) = (10x15x14)/(25x24x23)=2100/13800=21/138 = 7/46.
The probability of selecting girl = 10/25.
Next, Probability of selecting 1 boy = 15/24 (24 because 1 girl is already selected).
Next, the Probability of selecting another boy = 14/23 (23 because 1 boy and 1 girl were already selected) (14 because 1 boy was already selected before).
So,
P(E) = n(E)/n(s) = (10x15x14)/(25x24x23)=2100/13800=21/138 = 7/46.
(3)
Bhargavi said:
7 years ago
On which table 1050/2300 can be divisible?
(2)
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