A fair die is toss six times. Defined a random variable as an odd number that appeared. Find the probability that exactly four of the tosses showed an odd number.
prob (odd on a single toss) = 3/6 = 1/2
prob(not odd) = 1/2
prob that of 6 tosses, exactly 4 will show an odd
= C(6,4) (1/2)^4 (1/2)^2
= 15(1/2)^6
= 15/64
A poker hand is a hand made of five cards chosen from a standard deck of cards. Suppose a random
variable J gives the number of hearths in the hand.
a. P(J=0)
b. P(J=1)
c. P(J=2)
d. P(J=3)
e. P(J=4)
f. P(J=5)
g. Construct the probability mass function and its corresponding histogram.
To find the probability that exactly four of the tosses showed an odd number, we need to calculate the probability of each individual toss resulting in an odd number and the probability of each individual toss resulting in an even number.
A fair die has six possible outcomes, numbered 1 to 6. Out of these six outcomes, three are odd numbers (1, 3, 5) and three are even numbers (2, 4, 6).
Let's calculate the probabilities:
Probability of an odd number (P(odd)) = Number of odd outcomes / Total number of outcomes
P(odd) = 3/6 = 1/2
Probability of an even number (P(even)) = Number of even outcomes / Total number of outcomes
P(even) = 3/6 = 1/2
Now, we want to find the probability of exactly four tosses showing an odd number. For this, we'll use the binomial probability formula:
P(k) = (nCk) * p^k * (1 - p)^(n - k)
Where:
P(k) is the probability of getting k successes (k tosses showing an odd number).
n is the total number of tosses (6 in this case).
k is the number of successes we are interested in (4 in this case).
p is the probability of getting a success (probability of an odd number, which is 1/2).
(1 - p) is the probability of getting a failure (probability of an even number, which is also 1/2).
nCk represents the number of ways to choose k successes out of n tosses and is calculated as n! / (k! * (n - k)!).
Using the formula, let's substitute the values:
P(4) = (6C4) * (1/2)^4 * (1 - 1/2)^(6 - 4)
P(4) = (6C4) * (1/2)^4 * (1/2)^2
P(4) = (6! / (4! * (6 - 4)!)) * (1/2)^6
P(4) = (6! / (4! * 2!)) * (1/2)^6
P(4) = (6 * 5 * 4! / (4! * 2 * 1)) * (1/2)^6
P(4) = (6 * 5 / 2) * (1/2)^6
P(4) = (15 / 2) * (1/2)^6
P(4) = 15/64
Therefore, the probability that exactly four of the tosses show an odd number is 15/64.