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If a six-sided die(1-6) and a ten-sided die(1-10) are rolled simultaneously, then what is the probability of rolling both with a number greater than four

Sagot :

MrRoyal

Answer:

[tex]Pr = 0.20[/tex]

Step-by-step explanation:

Given

[tex]n_1 = 10[/tex] ---- 10 sided

[tex]n_2 = 6[/tex] --- 6 sided

Required

[tex]P(x_1,x_2 > 4)[/tex]

First, list out the sample space

[tex]S = \{(1, 1) (1, 2) (1, 3) (1, 4) (1, 5) (1, 6)(2, 1) (2, 2) (2, 3) (2, 4) (2, 5) (2, 6)[/tex]

[tex](3, 1) (3, 2) (3, 3) (3, 4) (3, 5) (3, 6) (4, 1) (4, 2) (4, 3) (4, 4) (4, 5) (4, 6)[/tex]

[tex](5, 1) (5, 2) (5, 3) (5, 4) (5, 5) (5, 6) (6, 1) (6, 2) (6, 3) (6, 4) (6, 5) (6, 6)[/tex]

[tex](7, 1) (7, 2) (7, 3) (7, 4) (7, 5) (7, 6) (8, 1) (8, 2) (8, 3) (8, 4) (8, 5) (8, 6)[/tex]

[tex](9, 1) (9, 2) (9, 3) (9, 4) (9, 5) (9, 6) (10, 1) (10, 2) (10, 3) (10, 4) (10, 5) (10, 6)\}[/tex]

[tex]n(S) = 60[/tex] --- total outcomes

The event that the outcome of both is greater than 4 is:

[tex]x_1,x_2> 4 = \{(5, 5) (5, 6)(6, 5) (6, 6)(7, 5) (7, 6) (8, 5) (8, 6)(9, 5) (9, 6) (10, 5) (10, 6)\}[/tex]

[tex]n(x_1,x_2>4) = 12[/tex]

So, the probability is:

[tex]Pr = \frac{n(x_1,x_2>4)}{n(S)}[/tex]

[tex]Pr = \frac{12}{60}[/tex]

[tex]Pr = 0.20[/tex]

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