Consider a set of N coins. If we toss each coin, each has two ways of coming dow

Question

Consider a set of N coins. If we toss each coin, each has two ways of coming down, H or T. Since the first coin can come down 2 ways, and the second coin can come down 2 ways, etc., the number of different ways (microstates) that the N coins can come down is 2 x 2 x ... (N times) = 2N. While this is interesting, this is not the number we want. Rather, we want to know if we choose a particular microstate (a given number of heads and tail) how many microstates correspond to that microstate. That is, how many different ways could you get a string of coin flips that came up with that particular number of heads and tails? A. For 4 coins, count explicitly how many different ways there are to get each of the following microstates: 4H, 0T 3H, 1T 2H, 2T 1H, 3T OH, 4T.

Explanation / Answer

When we toss we have total 2^4 = 16 possibilities. which are given by

HHHH      THHH

HHHT      THHT

HHTH      THTH

HHTT      THTT

HTHH      TTHH

HTHT      TTHT

HTTH      TTTH

HTTT       TTTT

If we can summerize this according to the number of heads then

Number of heads number of outcomes with N heads     probability to get N heads

0 1                                       1/16 = 0.0625

1 4                                       4/16 = 1/4 = 0.25

2 6                                      6/16 = 3/8 = 0.375

3 4                                      4/16 = 1/4 = 0.25

4 1                                      1/16 = 0.0625

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