I am working from home today and thus using a datacard for net connection. The connection is really slow. Hope the question posted below reaches the forum.
If you flip a coin five times, what's the probability you get heads exactly twice?
Is there any generalized way to do these kind of questions? I mean if I flip a coin 100 times, whats the probability that I get exactly 49 heads? Is there a specific way to do this?
If you flip a coin five times, what's the probability you get heads exactly twice?
Is there any generalized way to do these kind of questions? I mean if I flip a coin 100 times, whats the probability that I get exactly 49 heads? Is there a specific way to do this?
Kindly help.
hi rahul... when you flip a coin 5 times, total possible outcomes is 2^5 = 32 We want heads exactly twice..and there is no restriction to get heads on any particular flip.. hence out of 5 flips , we have to select any 2 flips to get heads..
Hence, this turns out to be pure selection and no arrangement. i.e we have to select 2 places out of 5 for heads, which can be done in 5C2 ways and we need exactly 3 tails in remaining 3 flips i. e 3C3 ways or just 1 way..
Hence, p(exactly 2 heads out of 5) = 5C2 / (2^5) = 10/32 = 5/16.
To generalize, if we want exactly r heads (or tails) out of n flips, P = nCr / (2^n) Hence, exactly 49 heads out of 100 flips , p = 100C49 / (2^100)
Ravikant wrote: hi guys...found this question tricky...need your help...am not able to arrive at an answer..
TWO couples and a single person are to be seated on 5 chairs such that no couple is seated next to each other. What is the probability of the above??
A] 1/5 B] 2/5 C] 3/5 D] 4/5 E] 1/20
thank a ton..
Had put up this question in the other forum I participate in. Had a very interesting approach to the solution.
Really liked it..... For your consumption!!
It's often easiest to break down the problem into cases- it's a technique we use all the time in counting problems. Let's call the single person S. S must be in the 1st, 2nd, 3rd, 4th or 5th seat:
-If S is in the first seat - 1 choice -then anyone else can be in the 2nd seat: 4 choices -in the 3rd seat, cannot be the partner of the person in the 2nd seat: 2 choices -in the 4th seat: must be the partner of the person in the 2nd seat- 1 choice -fifth seat: must be the partner of the person in the 3rd seat- 1 choice.
So if S is in the first seat, we have 1*4*2*1*1 = 8 arrangements. Notice we'll get the same answer when S is in the fifth seat.
It's just as fast to count the arrangements with S in the second/fourth seat (8 arrangements for each) and third seat (16 arrangements), which gives us 8+8+16+8+8 = 48 arrangements in total. So the answer must be 48/5! = 2/5.
bhavin wrote: marcel wrote: i have 1 more query...
In how many ways can one choose 6 cards from a normal deck of cards so as to have all suits present?
a. (13^4) x 48 x 47 b. (13^4) x 27 x 47 c. 48C6 d. 13^4 e. (13^4) x 48C6
pls help...
Take a look at this
Bhavin,
I have a doubt in this question.In the selection of last 2 cards u r selecting in individual ways ie 48 & 47 ways........whats the error if we choose last 2 cards from 48 cards so selection would be 48C2......
I have a doubt in this question.In the selection of last 2 cards u r selecting in individual ways ie 48 & 47 ways........whats the error if we choose last 2 cards from 48 cards so selection would be 48C2......
Hi Binish,
In the absence of Bhavin, let me try to explain the same to you. Would request Bhavin to clarify once he logs in whenever its possible for him.
Also would request other 'Quant' moderators to chip in with their suggestion.
To choose 2 cards from the rest of the 48 cards, the order is not important. We have already chosen 4 cards from the 4 main suites. The one major thing that restricted the choice was that at all suites should be represented. That has been done when the 4 cards were chosen.
Rest 2 cards can be picked from the rest 48 cards in any way. There is no restriction and no order. You can choose any card without restrain. It does not matter that the first should be Queen of Hearts and the next should be 2 of diamonds or any other combination.
Hence choose the next card with gay abandon from any of the 48 cards in 48 ways. One gone and one to go. Left -47 cards. Choose the last card from any of the 47 cards in 47 ways.
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