Aptitude - Probability - Discussion

@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.

The answer above is wrong. There are 3 ways to get 1 girl and 2 boys when choosing 3: (1)BBG, (2)BGB, (3)GBB. Probability of observing outcome (1) is 15/25*15/25*10/25=.6*.6*.4=.144.The probability of observing outcome (2) is 15/25*10/25*1/25=.6*.4*.6=.144. The probability of observing outcome (3) is 10/25*10/25*15/25=.4*.6*.6=.144.

Sum the three possible outcomes to get the probablity of the event .144+.144+.144=.432.

This is just a binomial probability where the probability of a success, say choosing a boy is 15/25=.6 and we want to know the probability of getting 2 boys in 3 chance, which is .432.

If they ask for same collections for example (a bag contains 6 white and 4 black balls.two balls are drawn at random.find the probability that they are of same color) here they are askin for same color in this case we have to use "+" i.e, 6c2+4c2

but here in a above problem they have tld us to select 1 girl out of 10 girls and 2 boys out of 15 boys.and here its not of same combination they are askin for both boy and gal.in this case we have to use "*" i.e, 10C1 x 15C2

Note tat: if its same color or combination go for "+"
If its different combination go for "*"

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.

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!.

@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.

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.

@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.

@Abhi we can do it by this way.

For girl : 10/25 (total students 10 + 15 = 25).
For 1st boy: 15/24 (1 girl is out so total is 24).
For 2nd boy: 14/23 (now total boys 14 and total students are 23).

[(10*15*14)/(25*24*23)]
=7/46.

But here three possibility so multiply by 3
1st 1g 1b 1b
2nd 1b 1g 1b
3rd 1b 1b 1g

So, answer = 21/46.

The answer is correct.
I have alternative solution.

There are 3 possible ways in picking 1 girl and 2 boys.

Girl , boy , boy
Boy , girl , boy
Boy , boy , girl

Gbb:
(10/25)*(15/24)*(14/23)=7/46.

Bgb:
(15/25)*(10/24)*(14/23)=7/46.

Bbg:
(15/25)*(14/24)*(10/23)=7/46.

Adding them all will give you 21/46.