Respuesta :
Answer:
The probability is 25/216. Approximately 0.116
Step-by-step explanation:
You need
- The first throw to fail (probability 1-1/6 = 5/6)
- The second throw to fail (probability 5/5)
- The third throw to be a success (probability 1/6)
Since each throw is independent of the others, we have to multiply all probabilities to obtain the total probability of the event. Thus, the probability of requiring 3 rolls until getting doubles is
5/6 * 5/6 * 1/6 = 25/216 = 0.115741
This problem can also be solved with sophisticated theory;
the random variable which counts the number of tries until the first success is a geometric distribution. The only parameter of the distribution is the probability of success p. If X is geometric with parameter p, then the probability of X being equal to k (in other words, requiring k tries for a success) is
P(X = k) = (1-p)^(k-1) * p
If p = 1/6, then
P(X = 2) = (1-1/6)^2*(1/6) = (5/6)²*1/6 = 25/216
The probability that it takes three rolls until the player rolls doubles is 0.11574 and this can be determined by using the given data.
Given :
- In a certain board game, a player rolls two fair six-sided dice until the player rolls doubles.
- The probability of rolling doubles with one roll of two fair six-sided dice is 1/6.
The probability of an event is nothing but a number between 0 and 1 where zero shows the impossibility and one shows the certainty.
First, determine the probability of rolls to obtain doubles in each roll.
Fails in the first roll:
[tex]\rm P = 1-\dfrac{1}{6} = \dfrac{5}{6}[/tex]
Fails in the second roll:
[tex]\rm P = 1-\dfrac{1}{6} = \dfrac{5}{6}[/tex]
In the third roll the player rolls doubles that is,
[tex]\rm P=\dfrac{1}{6}[/tex]
Now, the probability of requiring 3 rolls until getting doubles will be:
[tex]=\dfrac{5}{6}\times \dfrac{5}{6} \times \dfrac{1}{6}[/tex]
[tex]=\dfrac{25}{216}[/tex]
= 0.11574
For more information, refer to the link given below:
https://brainly.com/question/21586810
