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?
Now I have the answer and everything, however I do not understand the work.
chances of 1 two and 5 ones = 6 * 1/2^6 = 6/64 = 3/32 = 0.09375
If you roll the second one 6 times,
chances of 1 two and 5 ones = 6 * 2/3 * 1/3^5 = 12/(3*243) = 12/729 = 4/243 = 0.01646
So the probability it was the second die is
(4/243) / (4/243 + 3/32) = (0.01646 / 0.11021) = 0.14936 or about 15%
The part I don't understand it is the " 6 * 2/3 * 1/3^5" and "6 * 1/2^6", Everything else I understand.
Why is 6 multiplied by (1/2) to the power of 5? and for the second dice, why is 6 multiplied to 2/3 and 1/3 to the power of 5??
Thank you in advance
Well, let me try to break it down for you!
When you roll the first die six times, the probability of getting exactly one two-spot and five one-spots can be calculated as follows:
The probability of getting a two-spot on any one roll is 1/2, since there are two possible outcomes (one-spot or two-spot) and only one of them is a two-spot.
Since there are six rolls, we need to multiply the probabilities together, giving us (1/2) * (1/2) * (1/2) * (1/2) * (1/2) * (1/2).
This can also be written as (1/2)^6 or (1/2) raised to the power of 6.
The same logic applies to the second die. However, since this die has four two-spots and two one-spots, the probability of getting a two-spot on any one roll is 2/3 (four possible outcomes, two of which are two-spots).
Once again, we multiply this probability by itself six times, giving us (2/3) * (2/3) * (2/3) * (2/3) * (2/3) * (2/3).
This can also be written as (2/3)^6 or (2/3) raised to the power of 6.
I hope that clarifies things a bit! Let me know if you have any more questions.
To understand why 6 is multiplied by (1/2)^5 and why 6 is multiplied by 2/3 and (1/3)^5, we need to break down the reasoning behind these calculations.
First, let's look at the expression 6 * 1/2^6.
When rolling the first die six times, we are interested in the probability of getting exactly one two-spot and the remaining five rolls resulting in one-spots.
Here's how the calculation is done:
- The probability of getting a two-spot on each roll is 1/2 (because there are two faces out of six with a two-spot).
- Since we want to calculate the probability of getting one two-spot and five one-spots, we need to consider the specific order in which these outcomes occur. There are six possible positions for the two-spot (first roll, second roll, etc.).
- Multiplying the probability of getting a two-spot (1/2) by the number of possible positions (6), we get the desired probability: 6 * 1/2 = 6/2 = 3/32.
Similarly, let's analyze the expression 6 * 2/3 * 1/3^5.
When rolling the second die six times, we want to find the probability of getting exactly one two-spot and the remaining five rolls resulting in one-spots.
Here's how the calculation is done:
- The probability of getting a two-spot on each roll is 2/3 (because there are four faces out of six with a two-spot).
- However, the probability changes as we roll the die each time. After each roll, the number of remaining two-spots decreases by one.
- This means that for the second die, the probability of getting a two-spot is 2/3 on the first roll, then 1/3 on the second roll (as one of the two-spots has already been rolled), and so on. So, the probability of getting a two-spot on each subsequent roll decreases.
- Considering the specific order in which these outcomes occur, we once again have six possible positions for the one two-spot (first roll, second roll, etc.).
- Multiplying all these probabilities together (2/3 * 1/3 * 1/3 * 1/3 * 1/3 * 1/3) and then multiplying by the number of positions (6), we get the desired probability: 6 * 2/3 * (1/3)^5 = 12/729 = 4/243.
To calculate the probability that it is the die with four two-spots, we compare the probability of rolling exactly one two-spot with each die. The calculation (4/243) / (4/243 + 3/32) compares the probability of rolling one two-spot with the second die (4/243) to the total probability of rolling one two-spot with either the first or second die (4/243 + 3/32).
This division allows us to determine the ratio of the probability of rolling one two-spot with the second die to the total probability of rolling one two-spot with either die. In this case, the result is approximately 0.14936 or about 15%, which represents the probability that the die with four two-spots was chosen.
In order to understand the calculations, let's break down the probability calculations step by step.
For the first die (three faces with one-spot and three faces with two-spot), the calculation "6 * 1/2^6" represents the probability of getting one two-spot and five one-spots in any order when rolling the die six times.
To understand this, let's consider the possible outcomes when rolling the die six times. Each roll has two possible outcomes: either a one-spot or a two-spot. Since there are six rolls, there are a total of 2^6 = 64 possible outcomes.
Out of these 64 outcomes, only 6 of them have exactly one two-spot and five one-spots. The "6" in the calculation accounts for these 6 favorable outcomes. The "1/2^6" part accounts for the probability of each favorable outcome occurring.
Since each roll has a 1/2 (or 0.5) probability of landing on a two-spot (because there are 2 faces out of 6 that are two-spots), we raise it to the power of 6 to account for all six rolls.
So, the probability of getting one two-spot and five one-spots for the first die is 6 * 1/2^6 = 6/64 = 3/32 = 0.09375.
Now, let's move on to the second die (two faces with one-spot and four faces with two-spot). The calculation "6 * 2/3 * 1/3^5" represents the probability of getting one two-spot and five one-spots in any order when rolling the second die six times.
Similarly, there are 2^6 = 64 possible outcomes when rolling the die six times. Out of these, we need to find the number of favorable outcomes where exactly one two-spot and five one-spots occur.
For the second die, there are only 2 faces out of 6 that are two-spots. So, the probability of getting a two-spot on any given roll is 2/6 = 1/3. Since we need exactly one two-spot, we multiply by 2/3.
Moreover, we raise the probability of getting a one-spot (1/3) to the power of 5 because we want five one-spots to occur on the remaining five rolls.
Therefore, the probability of getting one two-spot and five one-spots for the second die is 6 * 2/3 * 1/3^5 = 12/729 = 4/243 = 0.01646.
To determine the probability that it is the die with four two-spots, we compare the probability of it being the second die (4/243) to the total probability of getting one two-spot (4/243) plus the probability of getting one two-spot with the first die (3/32).
Thus, the probability it was the second die is (4/243) / (4/243 + 3/32) = 0.14936 or approximately 15%.
Hope this explanation clarifies the calculations for you!