The ans to your ques would be 2/7 i think (can never be sure when its probability )

And now just adding the options to the ques that I had posted earlier :

What is the probability of an year having 53 Sundays ?

A. 5/28
B. 13/28
C. 1/7
D. 1/84
E. None of these

Well the answer is 5/28 but that is in the case when u dont consider the fact that 100, 200,300 are not leap years..in that case it will be very complex one..but if you neglect them and consider every fourth year as leap year than the answer is ..i will say 5/28

Three cards are drawn from a well shuffled pack of 52 cards. Find the probability that all the three cards are of different suits and there is at least one face card.

Three cards are drawn from a well shuffled pack of 52 cards. Find the probability that all the three cards are of different suits and there is at least one face card.

out of 52 cards 40 are number cards and 12 are face cards.

case1:- out of 3, 2 are number cards and 1 is face card
no of ways= 40C1*30C1*6C1=7200....( only 6 face cards are to be considerd to make all 3 cards of different colour)

case2:- out of 3, 1 is number card and 2 are face cards
no of ways= 40C1*9C1*6C1=2160

case3:- all 3 are face cards
no of ways= 12C1*9C1*6C1=648

Hence total no of ways=7200+2160+648=10008

no of ways of drawing 3 cards out of 52= 52C3=22100

hence required prob.=10008/22100=1502/5525, which is not in the options...........
where is the mistake i don't know. can anyone tell me my mistake.........???

Three cards are drawn from a well shuffled pack of 52 cards. Find the probability that all the three cards are of different suits and there is at least one face card.

Earlier I was getting sum weird answer. But then I solved by this method and got the rgt answer. Slowly I guess I will be back to solving such questions without much difficukty

Yes that is right answer. Even I got same. Probability is difficult. Sumtimes 10 different people get 10 different answer. Sometimes even we will get 2 diffrent answer when we do sum 2 diffrent times

Yes that is right answer. Even I got same. Probability is difficult. Sumtimes 10 different people get 10 different answer. Sometimes even we will get 2 diffrent answer when we do sum 2 diffrent times

thats the beauty of probability and thats why one needs to practice a lot in Prob

No of ways to select 3 cards from a pack of cards : 52C3

Ways to select 3 suites out of 4 (this is because all the 3 cards have to be of different suites) : 4C3

Now, P(to select at least 1 face card) =
P(to select any card of that suit) - P(any non-face card of that suit)

But since the 3 cards are drawn together, so we have

(13C1)^3 - (10C1)^3

Hence, the solution is :

4C3*[(13C1)^3 - (10C1)^3] / (52C3)

= 1197/5525 (phew !!!)

ok tushar u got the right answer with a right approach.
but still i m asking wat went wrong in my n sumit's method?
itz also very necessary to clear our doubts regarding the way we approach for any question if we don't get the correct answer.

ok tushar u got the right answer with a right approach.
but still i m asking wat went wrong in my n sumit's method?
itz also very necessary to clear our doubts regarding the way we approach for any question if we don't get the correct answer.

i got the flaw in my method as i was doing 1 - 4C3 * 1000/22100 here actually i am subracting the case of different suits as well it should be as mentioned by tushar

ok tushar u got the right answer with a right approach.
but still i m asking wat went wrong in my n sumit's method?
itz also very necessary to clear our doubts regarding the way we approach for any question if we don't get the correct answer.

Yes yes, I totally ralise that....actually am in office right now so thats why couldnt reply...its only now that I have got some time

Think sumit has already recognised his mistake and pointed out yours too...

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