For a fair coin tossed three times, the eight possible simple events are HHH, HHT, HTH, THH, HTT, THT, TTH, TTT. Let X = number of tails. Find the probability distribution for X by filling in the table showing each possible value of X along with the probability that value occurs. k=0,1,2,3 P(X=k)

k _____ P(x=k)

0 ....... C(3,0)(1/2^0(1/2^3 = 1/8
1 ....... C(3,1)(1/2^1(1/2)^2 = 3/8
2 ....... C(3,2)(1/2)^2(1/2)^1 = 3/8
3 ........ C(3,3)(1/2)^3(1/2)^0 = 1/8

notice the sum of those is 1, as it should be.

To find the probability distribution for X, we need to determine the probability of each possible value of X, where X represents the number of tails.

Let's calculate the probability for each value of X:

When X = 0 (no tails):
The only possible outcome is HHH.
P(X=0) = 1/8

When X = 1 (one tail):
There are three possible outcomes: HTH, THH, and HHT.
P(X=1) = 3/8

When X = 2 (two tails):
There are three possible outcomes: HTT, THT, and TTH.
P(X=2) = 3/8

When X = 3 (three tails):
The only possible outcome is TTT.
P(X=3) = 1/8

Now, let's fill in the table:

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| k | P(X=k) |
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| 0 | 1/8 |
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| 1 | 3/8 |
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| 2 | 3/8 |
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| 3 | 1/8 |
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This table represents the probability distribution for X, showing the possible values of X (0, 1, 2, and 3) along with the corresponding probabilities (1/8, 3/8, 3/8, and 1/8 respectively).

To find the probability distribution for the number of tails (X) when a fair coin is tossed three times, we can analyze each possible outcome and calculate the probability of each value of X.

There are a total of 8 possible outcomes when a fair coin is tossed three times. We can assign the value of X for each outcome based on the number of tails. Here is the table:

X | Outcome | k | P(X=k)
---|-----------------|---|--------
0 | HHH | 0 | P(X=0)
1 | HHT, HTH, THH | 1 | P(X=1)
2 | HTT, THT, TTH | 2 | P(X=2)
3 | TTT | 3 | P(X=3)

Since the coin is fair, each individual coin toss has a 50% chance of landing on heads (H) and tails (T).

To calculate the probability of each value of X, we can compute the number of outcomes that corresponds to each value and divide it by the total number of possible outcomes.

P(X=0) is the probability of having zero tails. From the table, we see that there is only one outcome (HHH) that has zero tails. Therefore, P(X=0) is 1/8 or 0.125.

P(X=1) is the probability of having one tail. From the table, we see that there are three outcomes (HHT, HTH, THH) that have one tail. Therefore, P(X=1) is 3/8 or 0.375.

P(X=2) is the probability of having two tails. From the table, we see that there are three outcomes (HTT, THT, TTH) that have two tails. Therefore, P(X=2) is 3/8 or 0.375.

P(X=3) is the probability of having three tails. From the table, we see that there is only one outcome (TTT) that has three tails. Therefore, P(X=3) is 1/8 or 0.125.

So, the probability distribution for X is:

X | P(X=k)
---|--------
0 | 0.125
1 | 0.375
2 | 0.375
3 | 0.125