A six faced is so constructed that the probability of getting an even number is twice the probability of getting an odd number. Find the probabilities that out of 5 trials
(1)all the five will give even numbers
(3)2 or 3 trails will give even numbers ....find by binomial distribution method...
Pr(even)=2/3
pr(odd)=1/3
a. Pr(all five even)=(2/3)^5
b. I dont understand...2 or 3 trials will give (how many) even numbers?
Yes I got it my answer thank you ...bobpursley
To find the probabilities using the binomial distribution method, we need to determine the probability of success (getting an even number) and the number of trials (5 in this case).
Let's let p represent the probability of getting an even number and q represent the probability of getting an odd number. Given that the probability of getting an even number is twice the probability of getting an odd number, we have the following relationship:
p = 2q
Since each die has six equally likely outcomes, the probability of getting an even number is 3/6 = 1/2, and the probability of getting an odd number is 3/6 = 1/2.
Now, let's calculate the probabilities using the binomial distribution method:
(1) Probability that all five trials will give even numbers:
In this case, we want to calculate the probability of getting 5 successes (even numbers) out of 5 trials. Using the binomial probability formula, we have:
P(X = k) = C(n,k) * p^k * q^(n-k)
P(X = 5) = C(5,5) * (1/2)^5 * (1/2)^0 (since q^0 = 1)
C(5,5) = 1 (combination of choosing all 5 trials out of 5)
P(X = 5) = 1 * (1/2)^5 * 1
P(X = 5) = 1/32
Therefore, the probability that all five trials will give even numbers is 1/32.
(2) Probability that 2 or 3 trials will give even numbers:
In this case, we want to calculate the probability of getting either 2 or 3 successes (even numbers) out of 5 trials. We can calculate this probability by adding the individual probabilities of getting 2 and 3 successes.
P(X = 2) = C(5,2) * (1/2)^2 * (1/2)^3 (using the binomial probability formula)
P(X = 2) = 10 * (1/4) * (1/8)
P(X = 2) = 10/32
P(X = 3) = C(5,3) * (1/2)^3 * (1/2)^2 (using the binomial probability formula)
P(X = 3) = 10 * (1/8) * (1/4)
P(X = 3) = 10/32
P(2 or 3) = P(X = 2) + P(X = 3)
P(2 or 3) = 10/32 + 10/32
P(2 or 3) = 20/32
Therefore, the probability that 2 or 3 trials will give even numbers is 20/32 or 5/8.
I hope this explanation helps! Let me know if you have any further questions.