Posted by **Elizabeth** on Saturday, December 5, 2009 at 4:29pm.

When a coin is tossed four times, sixteen equally likely outcomes are possible as shown below:

HHHH HHHT HHTH HHTT

HTHH HTHT HTTH HTTT

THHH THHT THTH THTT

TTHH TTHT TTTH TTTT

Let X denote the total number of tails obtained in the four tosses. Find the probability distribution of the random variable X. Leave your probabilities in fraction form.

- math -
**MathMate**, Sunday, December 6, 2009 at 11:07am
As you mentioned, there are 16 possible outcomes each with equal probability.

For X=0, there is only one case out of sixteen, namely HHHH. Therefore

X(0)=1/16.

For X=1, you will count the number of cases where T occurs only once. You should count 4 of such cases, therefore

X(1)=4/16=1/4

Repeat the calculation for X=2,3,4 and obtain the values of X(2), X(3), and X(4).

The sum of the values X(0) to X(5) should equal to 1.

- math -
**PsyDAG**, Sunday, December 6, 2009 at 11:10am
If there are 16 possible outcomes for the 4 tosses, the probability of getting 4 tails = 1/16, 3 tails = 4/16 = 1/4, 2 tails = ?, 1 tail = ?, no tails = ?.

I'll let you calculate the remaining fractions.

As a check, the sum of these fractions must = 1.

I hope this helps.

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