DSE 2010

[Swati Kanodia]

Can u please help me solve the following question:

Suppose  player 1 has 5 coins and player 2 has 4 coins. Both players toss all their coins and observe the number that come up heads. Assuming all the coins are fair, what is the probability that player 1 obtains more heads than player 2?

I solved the question assuming the following combinations : (1H, 0H), (2H, 0H) (2H, 1H) (3H, 0H) (3H, 1H) (3H , 2H) (4H, 0H) (4H, 1H) (4h, 2H) (4H, 3H) (5H, 0H) (5H, 1H) (5H, 2H) (5H, 3H) (5H, 4H)...But i am not getting the answer..:(((

[Amit Goyal]

You are right these are exactly the cases you need:
Pr((1H, 0H))= (C(5,1)(1/32))(C(4,0)1/16)
Pr((2H, 0H))= (C(5,2)(1/32))(C(4,0)1/16)
Pr((2H, 1H))= (C(5,2)(1/32))(C(4,1)1/16)
Pr((3H, 0H))= (C(5,3)(1/32))(C(4,0)1/16)
Pr((3H, 1H))= (C(5,3)(1/32))(C(4,1)1/16)
Pr((3H, 2H))= (C(5,3)(1/32))(C(4,2)1/16)
Pr((4H, 0H))= (C(5,4)(1/32))(C(4,0)1/16)
Pr((4H, 1H))= (C(5,4)(1/32))(C(4,1)1/16)
Pr((4h, 2H))= (C(5,4)(1/32))(C(4,2)1/16)
Pr((4H, 3H))= (C(5,4)(1/32))(C(4,3)1/16)
Pr((5H, 0H))= (C(5,5)(1/32))(C(4,0)1/16)
Pr((5H, 1H))= (C(5,5)(1/32))(C(4,1)1/16)
Pr((5H, 2H))= (C(5,5)(1/32))(C(4,2)1/16)
Pr((5H, 3H))= (C(5,5)(1/32))(C(4,3)1/16)
Pr((5H, 4H))= (C(5,5)(1/32))(C(4,4)1/16)
Sum them up and you will get 0.5

[Swati Kanodia]

Thank you sir for your quick response...Actually i forgot to apply the combinations rule..Thank you...:))

[riddhima]

Hello Sir
i have solved the question in the manner you have just prescribed above, but my answer came out to be 0.5976 ans  not 0.5, the next closest choice was 5/9

My final answer after summing all them up is as follows

1/32*16( 5+10+10+5+1+40+40+20+4+60+60+6+40+4+1) = 306/512

Please clarify if and where I am wrong!

Thanks!

[duck]

Hi Ridhimma.. :)

You've done some calculation error. Please check.

=1/32*16 [5+10+40+10+40+60+5+20+30+20+1+4+6+4+1]
= 256/32*16
= 1/2