Question

can anyone help me on how to find the probability of getting a head by tossing N coins out of which half of them are biased (only tail falls)...?

Answer 1

N/2 (1/2)

N/2 (0) for the others

Answer 2

looks to me that out of N coins tossed, the prob of getting heads is N/2 (1/2) of them

Answer 3

but i have a tendency to read these wrong ....

Answer 4

spose N=2 coins

HT
TT
TT
TT

would be the likely outcomes with a biased coin

so 1/4 seems right to me

Answer 5

@amistre64 if N/2 *(1/2) is probability then consider there are 10 coins according to you we will probability as 2.5. we cant get probability greater than 1.

Answer 6

maybe N/2*(1/2)^(N/2) ???

P(Hu)*P(Tu)*P(Tu)* ...(unbiased times) * P(Tb)*P(Tb)*P(Tb)*.... (biased times)

.5*.5*.5* ... *1*1*1*....

thats the underlieing idea i have, but as a siad, im prolly wrong :)

Answer 7

I would imagine that there are N/2 ways to get 1H in that setup

Answer 8

if we need to find the probability of a head we can take like N/2C1 multiplied by probability of getting head(1/2) whole divided by N C1....
will that be right

Answer 9

since the probability of heads is .5; and the probability of tails is .5 ... ideally

the prob of getting: Hu and Tu and Tu ... and Tb and Tb and Tb ....
equals: .5 * .5 * .5 ... * 1 * 1 * 1

the number of times we can get an H is N/2 times

the total number of outcomes i believe is \[\frac{N!}{dunno}\]hmm

Answer 10

im sure Zarkon or Satellite would be better equipped to determine a suitable result tho