A specially made pair of dice has only one- and two-spots on the
faces. One of the dice has three faces with one-spot and three faces with two-spot.
The other die has two faces with one-spot and four faces with a two-spot. One of the
dice is selected at random and then rolled six times. If a two-spot shows up only once,
what is the probability that it is the die with four two-spots?
First one, roll 6 times
p is a two = 1/2
p not a two = 1/2
p a two one out six rolls
= (6,1) (6!)/[(1)!(5!)] (1/2)^1(1/2)^5
= 6 (1/2)^6
Second one
p is a two = 2/3
p not a two = 1/3
p a two one out of 6 rolls
= 6 (2/3)^1 (1/3)^5
= 6 (2)/3^6
P second one / p first one = 2/3^6 /1/2^6
= 2 (2/3)^6
.1756
Now IN THIS EXPERIMENT where all other outcomes are thrown outthe probability that it is the first + the probability that it is the second = 1
so
P (first) + .1756 P( first) = 1
P(first) = .851
so
P(second) = .149
1/36
To solve this problem, we need to apply Bayes' Theorem. Let's define the events:
A: The die with four two-spots is selected.
B: A two-spot shows up only once.
We want to find the probability of event A given that event B has occurred, denoted by P(A|B).
We can calculate this probability using Bayes' Theorem:
P(A|B) = (P(B|A) * P(A)) / P(B)
Now let's calculate each part of the equation step-by-step:
Step 1: Calculating P(B|A) - the probability of event B given that event A has occurred.
Since the die with four two-spots will give a two-spot exactly once out of six rolls, the probability is given by the binomial distribution: P(B|A) = C(6, 1) * (1/2)^1 * (1/2)^5 = 6 * (1/2) * (1/2)^5 = 6 * (1/2)^6
Step 2: Calculating P(A) - the probability of event A (selecting the die with four two-spots) without considering the occurrence of event B.
Since one of the dice is selected randomly, the probability of selecting the die with four two-spots is 1/2.
Step 3: Calculating P(B) - the probability of event B (a two-spot showing up only once) without considering event A.
We can calculate this probability by summing the probabilities of getting a two-spot once with each die:
P(B) = P(B|A) * P(A) + P(B|Ac) * P(Ac)
where Ac denotes the complement of event A.
Using similar calculations as in Step 1, we have:
P(B|Ac) = C(6, 1) * (2/3) * (1/3)^5 = 6 * (2/3) * (1/3)^5 = 12 * (1/3)^6
Therefore:
P(B) = P(B|A) * P(A) + P(B|Ac) * P(Ac) = (6 * (1/2)^6 * (1/2)) + (12 * (1/3)^6 * (1/2)) = (6/2^7) + (12/3^7)
Step 4: Calculating P(A|B) - the probability of event A given that event B has occurred.
Using the values calculated in Steps 1, 2, and 3:
P(A|B) = (P(B|A) * P(A)) / P(B) = [(6 * (1/2)^6 * (1/2)) / (6/2^7 + 12/3^7)]
Now, let's simplify the expression and calculate the exact value.
To calculate the probability, we will use conditional probability. Let's break down the problem step by step.
Step 1:
There are two possibilities for the selected die: the die with three one-spots and three two-spots or the die with two one-spots and four two-spots. Since one two-spot showed up, we want to find the probability that it is the die with four two-spots.
Step 2:
Let's define our events:
A: Selecting the die with four two-spots.
B: Rolling one two-spot.
Step 3:
We want to find P(A | B), the probability of event A occurring given that event B occurred. This is our desired probability.
Step 4:
We can use Bayes' theorem to solve this problem. Bayes' theorem states:
P(A | B) = (P(B | A) * P(A)) / P(B)
Where:
P(A) is the probability of event A occurring.
P(B) is the probability of event B occurring.
P(B | A) is the probability of event B occurring given that event A occurred.
Step 5:
Let's calculate each of these probabilities:
P(A) = the probability of selecting the die with four two-spots out of the two dice, which is 1/2 since each die is equally likely to be chosen.
P(B | A) = the probability of rolling one two-spot given that the die with four two-spots was selected. Since this die has four two-spots out of six faces, the probability of rolling one two-spot is 4/6.
P(B) = the probability of rolling one two-spot, regardless of which die was chosen. We can calculate this by considering the probabilities of rolling one two-spot for each die and then taking the weighted average based on the probabilities of selecting each die.
For the first die (three one-spots and three two-spots), the probability of rolling one two-spot is 3/6.
For the second die (two one-spots and four two-spots), the probability of rolling one two-spot is 2/6.
The overall probability of rolling one two-spot, regardless of which die was chosen, can then be calculated as (1/2) * (3/6) + (1/2) * (2/6) = 5/12.
Now we have all the necessary probabilities to use Bayes' theorem:
P(A | B) = (P(B | A) * P(A)) / P(B) = (4/6 * 1/2) / (5/12) = (2/3) / (5/12) = 8/15.
Therefore, the probability that the die with four two-spots was chosen given that one two-spot showed up is 8/15.