### Discussion :: Probability - General Questions (Q.No.7)

Prabhu said: (May 13, 2011) | |

(10C1 x 15C2) why they are multiplying this ? They should add the value know .. 10 c1 + 15 c2 |

Priya said: (May 20, 2011) | |

@prabhu. Even I have the same doubt. Can some intellectual person answer this? |

Gnana said: (May 23, 2011) | |

I too have same doubt. Can any one clarify this, soon. |

Mahesh Babu said: (Jun 17, 2011) | |

Because he asked 10 girls (and) 15 boys. |

Sanjana said: (Jun 26, 2011) | |

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 "*" |

Prabhat said: (Aug 28, 2011) | |

Nice explanation sanjana. |

Abhi said: (Aug 29, 2011) | |

I solved this problem as (10/25)*(15/24)*(14/23) because 1st select 1 girl from 10 out of total 25 students, then 1 boy from 15 out of remaining 24 then again 1 boy from 14 boys and total remainig 23 students.... i got 7/46 as my ans..... Can anyone please tell me whats wrong? please.. |

Kruvy said: (Sep 28, 2011) | |

You got the Point @Abhi .. :-) However I cannot help you in this. As I have no clue about is this a right method and robust to solve same kind of problem...??! Help us. And I Like this site. Appreciate it @IndiaBix. |

Swetha said: (Nov 5, 2011) | |

Thanks sanjana. |

Al Ford said: (Nov 10, 2011) | |

I have a similar problem. Can anyone explain how to answer this: A class contains 7 women and 7 men. Suppose we choose three random students. How many ways can we choose exactly 1 woman? |

Raja said: (Jan 1, 2012) | |

Very good sanjana. |

Superprofessor Extraordinaire said: (Apr 27, 2012) | |

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. |

Shyam said: (May 30, 2012) | |

@Abhi. I have the same problem. I am getting 7/46 But by formula the answer is given as 21/46. I solved it by (10/25) * (15/24) * (14/23). Please anyone explain on this query. |

Shyam said: (May 30, 2012) | |

@Abhi. I have got the solution for our problem. We are computing the probability of just one combination. i.e we need to do P(1g,1b,1b)+P(1b,1g,1b)+P(1b,1b,1g). we found only the first combination. :) Hope you understood. |

Annie said: (Aug 19, 2012) | |

@Abhi Because we do not care about the order in which we pick the boys and girls. Also, if you picked the boys and girls in a different order than the one you used, wouldn't it be a different answer? So we don't solve it like x/25 x y/24 x z/23 |

Ankit Patel said: (Feb 10, 2013) | |

@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. |

Nemi said: (Feb 26, 2013) | |

If there are 5 men and some women in a group. If the probability to select two Girls out of all is 3/20 then find the number of Girls in a group. Give answer if any one know. |

Ajay said: (Aug 13, 2013) | |

We can't take (10/25) (15/24) (14/23). , because all 3 students are selected at once, not one after another. |

Jay Lee Lord said: (Jul 4, 2014) | |

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. |

Dhruvil said: (Aug 19, 2014) | |

Probability = (15*14+10/25*24*23). = (220/25*24*23) = (11*2*10/25*24*23) = (11/5*6*23) = (11/690). |

Priya said: (Aug 29, 2014) | |

1050/2300 = 21/46. |

Yugesh said: (Nov 9, 2014) | |

Why do you multiply 24x25x23? |

Usman said: (Nov 23, 2014) | |

It is very simple in the probability theorem and means *(multiplication)n or means +(addition). So in the above question they asked to find out probability of selected students are one girl and two boy. So 10c1*15c2/25c3 = 1050/2300 = 21|46. |

Naresh said: (May 7, 2015) | |

"And" is asked in problem -> Multiplication. "Or" is asked in problem -> Addition. |

Asc said: (Jul 1, 2015) | |

@Abhi. You got this wrong when you took only one girl as the first solution. You could also have one boy and another boy as the first student chosen i.e 1. |

John said: (Aug 19, 2015) | |

Whether you choose BBG, BGB or GBB you have the same numerators of 15, 14 and 10 AND the same denominators of 25, 24 and 23. That works out as (15x14x10 = 2100) and (25x24x23 = 13800) which factors down to 7/46. I think I hope! |

Chandni said: (Oct 16, 2015) | |

Can anybody tell me why are they using 25C3? |

Mary-Ann said: (Jan 3, 2016) | |

I can't understand someone help. |

Naresh Rajpurohit said: (Jan 31, 2016) | |

25 students and 3 select boys and girl. So number ways of selecting 3 students out of 25. So use 25C3. And Probability = 10C1 (10 girl in choose any one) * 15C2 (15 boys in choose any two)/25C3 (All students in select 1 girl and 2 boys). |

Azly said: (Jun 19, 2016) | |

Thank you! @Sanjana. |

Kathir Lee said: (Jul 12, 2016) | |

@ Sanjana. Thank you for the explanation. |

Anony said: (Jul 29, 2016) | |

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. |

Person said: (Aug 9, 2016) | |

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. |

Archana Mahour said: (Sep 27, 2016) | |

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? |

Jahid said: (Oct 15, 2016) | |

@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. |

Jitender said: (Dec 7, 2016) | |

@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. |

Annieval said: (Jan 9, 2017) | |

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: (Feb 5, 2017) | |

It is simple. Only using the equation (n* (n-1) )/r!. We will get the solution. |

Lesly said: (Mar 27, 2017) | |

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. |

Kuldeep said: (Sep 1, 2017) | |

Thanks @Sanjana. |

Shikha said: (Sep 14, 2017) | |

I am not able to understand this answer please some one clear my doubt. |

Sanu said: (Sep 26, 2017) | |

We don't have to use nCr here it should be nPr. We are only selecting the girls and not arranging, so no need of using combination here. |

Deel Khus said: (Dec 11, 2017) | |

@Prabhu. They are multiplied bc. The rule of calculating probability is that if getting a or b then we must be added the probability and if getting a and b then its probability must be multiplied. |

Shekhar Gupta said: (Feb 13, 2018) | |

To select boy and girls are mutually independent events. So they are multiplied. |

Bhargavi said: (Jul 23, 2018) | |

On which table 1050/2300 can be divisible? |

Mahendra said: (Oct 27, 2018) | |

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. |

Jeevan said: (Jan 7, 2019) | |

@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. |

K Pradeep Kumar Reddy said: (Dec 31, 2019) | |

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. |

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