Alok Jha wrote:clearly statement 1 is not sufficient

for 2 . probablity of 1m & 1f is 21/50 = 7*3/5*5*2

=> 7/10 * 3/5 thus by this we can find the probablity

2 is sufficient

Alok can you just elaborate on this alittle more. I get that 1 is insufficient just by looking at it. But how does breaking down 21/50 into prime numbers allow us to solve the question? I've never seen a question solved this way so I'm curious, why does "7/10 * 3/5" mean we can solve it?

If two students are chosen at random with replacement from a certain class, what is the probability that two male students or two female students are selected?

1. There are 50 male students in the class.
2. The probability of selecting one male and one female student is 21/50.

Let's assume we have x males & y females in the class.

Number of ways in which we can select two students with replacement is
(x + y)C1 * (x + y)C1 = (x + y)^2

Number of of ways in which we can select 2 males or 2 females with replacement is

x^2/(x + y)^2 + y^2 /(x + y)^2

Now consider the options -

1. There are 50 male students in the class. -- Not sufficient to answer the question
2. The probability of selecting one male and one female student is 21/50.

Number of ways one male and one female can be selected
xC1/(x + y)^2 * yC1/(x + y)^2 = xy/(x+y)^2=21/50

Now to answer our question ,Number of of ways in which we can select 2 males or 2 females with replacement , we need to find the value of

x^2/(x + y)^2 + y^2 /(x + y)^2
=> x^2+y^2/(x + y)^2
=>((x + y)^2-2xy)/(x + y)^2 = 1- 2xy/(x + y)^2= 1-21/25=4/25


Hence B

arora007 wrote:If two students are chosen at random with replacement from a certain class, what is the probability that two male students or two female students are selected?
1) There are 50 male students in the class.
2) The probability of selecting one male and one female student is 21/50.

OA is B, but can somebody calculate and show the exact probability
This question is from papgust's 300 questions!

Ordinarily, for a question that almost everyone got right, I wouldn't think about posting. I saw almost everyone posting choose B as the correct answer. However, I saw more than half the people making a calculation mistake that might have caused them dearly in a PS question. Hence, I decided to write this post.

First of all, the easiest way to approach the question.

1) There are 50 male students in the class.
If we only know the number of male students, we can't calculate the probability of female students. Insufficient.


2) The probability of selecting one male and one female student is 21/50.

Now, when we choose two students, we can have only 3 cases:
1. Two males.
2. Two females
3. One male and one female.

Barring these three, there are no other cases. Total probability equals 1. Hence we can find probability of selecting two males or two females. Sufficient. And we have to do no calculations.

Now let's do the calculations anyway:-
Total probability of 3 cases = 1
We can find probability of selecting two males or two females = 1 - (probability of selecting one male and one female) = 1 - 21/50 = 29/50. Sufficient.
So the correct OA is, as most people got, B.

Now for the important conceptual part. Some people calculated that "with replacement", the required probability = 4/25. This answer is incorrect. First of all, the total probability rule of all possible cases equaling 1 is always applicable, regardless of whether students are chosen "with" or "without replacement". The final answer with/without replacement in both cases will be 29/50.

This is where a number of people made the mistake:

The probability of choosing one female and one male "with replacement" is NOT equal to mf/(m+f)^2.
The probability of choosing one female and one male correctly equals 2mf/(m+f)^2.

The two comes from the fact that, when we calculate probability in this way, we are essentially calculating permutational probability. So when we consider this way, we need to find all the possible arrangements.

For the one male and one female case, we have 2 possible arrangements:
1. Choose male first, choose female second.
2. Choose female first, choose male second.
Each of the above cases has the probability = mf/(m+f)^2 giving us the total probability of 2mf/(m+f)^2.

Hence in this case probability of selecting either two males or two females
= m*m/(m+f)^2 + f*f/(m+f)^2
=(m^2+f^2)/(m+f)^2
=(m^2+f^2+2mf-2mf)/(m+f)^2
=((m+f)^2-2mf)/(m+f)^2
= (m+f)^2/(m+f)^2 -2mf/(m+f)^2
= 1 - 21/50
= 29/50.

So we arrive at the same answer as we got earlier from the total probability rule.

Key takeaways from this post:

1. The total probability rule applies regardless of whether the things are selected with or without replacement.
2. When doing probability questions, always be considerate of whether your way is arrangement or selection based.

I hope the two points above helps you guys get the right answer often.

Cheers!

P = number of probable outcomes/number of possible outcomes
P (M & F) = P (M) * P (F)
Let total number be x
So P( M & F) = 50/x * x-50/x
Hence, 1 is not sufficient
So input statement 2
21/50 = 50/x * x-50/x
We can solve for x, so both are sufficient together- C

Does replacement indicate that the new student is the exact same (I.e gender) as the student being replaced?

If not, wouldn't this change the numerator? (m)(m-1)? I know this leaves the denominator (m+w) unchanged.

1) Insufficient

2) sufficient

Ans: B

E in answer

P(M).P(F)=21/50
We need P(M)^2 + P(F)^2
P(M)+P(F)=1
P(M)^2+P(F)^2+2*P(M)P(F)=1
Hence P(M)^2 + P(F)^2=4/25

P(MF or FM) = 21/50

P(MF or FM or MM or FF) = P(MF or FM) + P(MM or FF) = 1

Therefore P(MM or FF) = 1 - 21/50 = 29/50

viveksaraswat26 wrote:P(M).P(F)=21/50
We need P(M)^2 + P(F)^2
P(M)+P(F)=1
P(M)^2+P(F)^2+2*P(M)P(F)=1
Hence P(M)^2 + P(F)^2=4/25

I see what you've done here, but there are also 2 ways of getting a male and a female: MF or FM

So P(M).P(F) + P(F).P(M) = 21/50

Now you can work out the surplus to make 1 (as you did above)

I hope this helps.

Duplicate message. Please ignore.

Duplicate message. Please ignore.

Duplicate message. Please ignore.

Mathsbuddy wrote:

viveksaraswat26 wrote:P(M).P(F)=21/50
We need P(M)^2 + P(F)^2
P(M)+P(F)=1
P(M)^2+P(F)^2+2*P(M)P(F)=1
Hence P(M)^2 + P(F)^2=4/25

I see what you've done here, but there are also 2 ways of getting a male and a female: MF or FM

So P(M).P(F) + P(F).P(M) = 21/50

Now you can work out the surplus to make 1 (as you did above)

I hope this helps.

Yeah! It really helped thrice.

tnkippen wrote:Does replacement indicate that the new student is the exact same (I.e gender) as the student being replaced?

If not, wouldn't this change the numerator? (m)(m-1)? I know this leaves the denominator (m+w) unchanged.

I believe it's intended to mean that the student is not removed.

Example: There are 3 blue balls and 2 red balls in a bag.
If you take a blue ball and not replace it into the bag, the second time you take a ball, the probability will be different. Replacing the ball would keep the probabilitiers the same.

I hope this helps.