Question #5958

find the probability distribution for the number of fails that occur when 4 coins are tossed?

Expert's answer

Let n be the number of fails.

Here are all possible results:

0 fails:

0 0 0 0

(1 possibility)

1 fail:

1 0 0 0

0 1 0 0

0 0 1 0

0 0 0 1

(4 possibilities)

2 fails:

1 1 0 0

1 0 1 0

1 0 0 1

0 1 1 0

0 1 0 1

0 0 1 1

(6 possibilities)

3 fails:

0 1 1 1

1 0 1 1

1 1 0 1

1 1 1 0

(4 possibilities)

4 fails:

1 1 1 1

(1 possibility)

Wee see that there are 1+4+6+4+1 = 16 possible rusults. So,

P(n=0) = 1/16,

P(n=1) = 4/16 = 1/4,

P(n=2) = 6/16 = 3/8,

P(n=3) = 1/4,

P(n=4) = 1/16.

Here are all possible results:

0 fails:

0 0 0 0

(1 possibility)

1 fail:

1 0 0 0

0 1 0 0

0 0 1 0

0 0 0 1

(4 possibilities)

2 fails:

1 1 0 0

1 0 1 0

1 0 0 1

0 1 1 0

0 1 0 1

0 0 1 1

(6 possibilities)

3 fails:

0 1 1 1

1 0 1 1

1 1 0 1

1 1 1 0

(4 possibilities)

4 fails:

1 1 1 1

(1 possibility)

Wee see that there are 1+4+6+4+1 = 16 possible rusults. So,

P(n=0) = 1/16,

P(n=1) = 4/16 = 1/4,

P(n=2) = 6/16 = 3/8,

P(n=3) = 1/4,

P(n=4) = 1/16.

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