Respuesta :
The probability of rolling a total of 7 on the two dice is 8/63
Probability:
In Mathematics, Probability is a branch that deals with the occurrence of possible outcomes in a random event or experiment. It is also defined as the ratio of the number of favorable outcomes and the total number of outcomes.
The probability of an event is denoted by P(E). And the probability of all the events in the sample space will equal to 1.
Probability P(E) = No of favorable outcomes /Total No of outcomes
Here we have,
The probabilities of rolling 1, 2, 3, 4, 5, and 6, on each die, are in the ratio 1: 2: 3: 4: 5: 6
Let x, 2x, 3x, 4x, 5x, and 6x are the probabilities of rolling 1, 2, 3, 4, 5, and 6, on each die [ since they are in ratio 1: 2: 3: 4: 5: 6 ]
As we know the total events in a probability = 1
=> x + 2x + 3x + 4x + 5x + 6x = 1
=> 21 x = 1
=> x = 1/21
From the above calculations,
The probabilities of rolling 1, 2, 3, 4, 5, and 6 are [tex]\frac{1}{21} , \frac{2}{21}, \frac{3}{21}, \frac{4}{21}, \frac{5}{21}, \frac{6}{21}[/tex] respectively.
The possible combinations of two dies that can give a total of 7 are
(1, 6) (2, 5) (3, 4) (4, 3) (5, 2) (6, 1)
Therefore, the probability of rolling a total of 7 on the two dice
= the sum of the probabilities of rolling each combination
= [tex]\frac{1}{21} . \frac{6}{21} + \frac{2}{21} .\frac{5}{21} + \frac{3}{21} . \frac{4}{21} + \frac{4}{21} . \frac{3}{21} + \frac{5}{21} .\frac{2}{21} + \frac{6}{21} .\frac{1}{21}[/tex]
= [tex]\frac{6}{441} + \frac{10}{441} + \frac{12}{441} + \frac{12}{441} + \frac{10}{441} + \frac{6}{441}[/tex]
= [tex]\frac{56}{441}[/tex]
= [tex]\frac{8}{63}[/tex]
Learn more about Probability at
https://brainly.com/question/13740302
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The complete question is
For a particular peculiar pair of dice, the probabilities of rolling 1, 2, 3, 4, 5, and 6, on each die are in the ratio 1:2:3:4:5:6. What is the probability of rolling a total of 7 on the two dice
