a fair coin is flipped 5 times the random variable is x is defined to be the number of heads that are observed identify the probability mass function of the random variable x.
x P(x)
To find the probability mass function (PMF) of the random variable x, which represents the number of heads observed when flipping a fair coin 5 times, we need to determine the probability of each possible outcome.
In this case, since we are flipping a fair coin, there are two possible outcomes for each flip - heads (H) or tails (T).
The number of ways we can get x heads out of 5 flips can be represented using the binomial coefficient formula:
C(n, k) = n! / (k!(n-k)!)
Where:
n = total number of flips
k = number of heads we want to observe
Let's calculate the probability for each value of x:
For x = 0:
P(0) = C(5, 0) * (0.5)^0 * (0.5)^(5-0) = 1 * 1 * 0.03125 = 0.03125
For x = 1:
P(1) = C(5, 1) * (0.5)^1 * (0.5)^(5-1) = 5 * 0.5 * 0.03125 = 0.15625
For x = 2:
P(2) = C(5, 2) * (0.5)^2 * (0.5)^(5-2) = 10 * 0.25 * 0.03125 = 0.15625
For x = 3:
P(3) = C(5, 3) * (0.5)^3 * (0.5)^(5-3) = 10 * 0.125 * 0.03125 = 0.09375
For x = 4:
P(4) = C(5, 4) * (0.5)^4 * (0.5)^(5-4) = 5 * 0.0625 * 0.03125 = 0.03125
For x = 5:
P(5) = C(5, 5) * (0.5)^5 * (0.5)^(5-5) = 1 * 0.03125 * 1 = 0.03125
Now we can summarize the PMF:
x P(x)
0 0.03125
1 0.15625
2 0.15625
3 0.09375
4 0.03125
5 0.03125
Note: The sum of all probabilities in the PMF should be equal to 1.
To identify the probability mass function (PMF) of the random variable x, which represents the number of heads observed when a fair coin is flipped 5 times, we need to find the probability of each possible outcome.
The number of possible outcomes when flipping a fair coin 5 times is 2^5 = 32. These outcomes can be represented by sequences of five coin flips, with H representing a heads outcome and T representing a tails outcome.
Let's calculate the probability for each possible value of x:
x = 0 (no heads):
The sequence can only be TTTTT, which has a probability of (1/2)^5 = 1/32.
x = 1 (one head):
There are five possible sequences: HTTTT, THTTT, TTHTT, TTTHT, and TTTTH. Each sequence has a probability of (1/2)^5 = 1/32. Therefore, the total probability is 5 * 1/32 = 5/32.
x = 2 (two heads):
There are ten possible sequences: HHTTT, HTHTT, HTTHT, HTTTH, THHTT, THTHT, THTTH, TTHHT, TTHTH, and TTTHH. Each sequence has a probability of (1/2)^5 = 1/32. Therefore, the total probability is 10 * 1/32 = 10/32 = 5/16.
x = 3 (three heads):
There are ten possible sequences: HHHTT, HHTHT, HHTTH, HTHHT, HTHTH, HTTHH, THHHT, THHTH, THTHH, and TTHHH. Each sequence has a probability of (1/2)^5 = 1/32. Therefore, the total probability is 10 * 1/32 = 10/32 = 5/16.
x = 4 (four heads):
There are five possible sequences: HHHHT, HHHTH, HHTHH, HTHHH, and THHHH. Each sequence has a probability of (1/2)^5 = 1/32. Therefore, the total probability is 5 * 1/32 = 5/32.
x = 5 (five heads):
The sequence can only be HHHHH, which has a probability of (1/2)^5 = 1/32.
Now we have calculated the probability for each possible value of x:
x P(x)
0 1/32
1 5/32
2 5/16
3 5/16
4 5/32
5 1/32
Therefore, the probability mass function (PMF) for the random variable x, which represents the number of heads observed when a fair coin is flipped 5 times, is:
x P(x)
0 1/32
1 5/32
2 5/16
3 5/16
4 5/32
5 1/32