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2 years ago

PLEASE HELP ME OUT DOWN BELOW!!!

Mathematics

1 answer:

Svetllana [295]2 years ago

4 0

Answer:

I believe the answer is B.

Step-by-step explanation:

If you look at the graph points for the last three hours it continues downword at the same rate.

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Consider the following game, played with three standard six-sided dice. If the player ends with all three dice showing the same

ddd [48]

Answer:

$P(A) = \frac{1}{36}$

$P(B) = \frac{15}{36}$

$P(U) = \frac{11}{36}$

Step-by-step explanation:

From the question we are told that

The number of dice is $n = 3$

Generally for all three dice to show the same number the second and the third dice must have the same outcome as the outcome of the first dice.

This means that the number of outcome the first die can have is 6 (6 sides), the number of outcome which the second and the third dice can have is 1 (they must match the first )

So the probability that all three dice show the same number on the first roll is mathematically represented as

$P(A) = \frac{6}{6} * \frac{1}{6} * \frac{1}{6}$

=> $P(A) = \frac{1}{36}$

Generally for two of the three dice show the same number after the first roll then the number of outcome for one of the dice would be is 6 , the number of outcome for another one must be 1(i.e it must match the first ) , the number of outcome for the remaining one must be 5 (i.e it can show any of the remaining 5 sides which the first and second dice are not showing )

Now the number of ways of selecting this 2 dice that show the same number from the 3 dice is mathematically represented as

$N = ^3C_2$

Here C stands for combination

$N = ^3C_2 = 6$

So the probability that exactly two of the three dice show the same number after the first roll is mathematically represented as

$P(B) = N \frac{1}{6} * \frac{5}{6}$

=> $P(B) = \frac{15}{36}$

Generally from the question we are told that if two dice match, the player re-rolls the die that does not match.

Now the probability that the die that did not match the first time will match the second time is

$P(E ) = \frac{1}{6}$

Generally if that one die does not show the same number in the second round , the probability that it will match in the third round is

$P(R) = \frac{5}{6} * \frac{1}{6}$

Generally the probability that he wins (i.e when all three are showing the same number ) and exactly two is showing the same number is mathematically represented as

$P(K) = P(E)* P(B) + P(R)* P(B)$

=> $P(K) = \frac{1}{6} * \frac{15}{36} + \frac{5}{6} * \frac{1}{6} * \frac{15}{36}$

=> $P(K)= \frac{165}{1296}$

So the probability that the player wins, conditioned on exactly two of the three dice showing the same number after the first roll is mathematically represented as

$P(U) = \frac{\frac{165}{1296}}{\frac{15}{36} }$