Question

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

Answer 1

Answer:

$Pr = 0.20$

Step-by-step explanation:

Given

$n_1 = 10$ ---- 10 sided

$n_2 = 6$ --- 6 sided

Required

$P(x_1,x_2 > 4)$

First, list out the sample space

$S = \{(1, 1) (1, 2) (1, 3) (1, 4) (1, 5) (1, 6)(2, 1) (2, 2) (2, 3) (2, 4) (2, 5) (2, 6)$

$(3, 1) (3, 2) (3, 3) (3, 4) (3, 5) (3, 6) (4, 1) (4, 2) (4, 3) (4, 4) (4, 5) (4, 6)$

$(5, 1) (5, 2) (5, 3) (5, 4) (5, 5) (5, 6) (6, 1) (6, 2) (6, 3) (6, 4) (6, 5) (6, 6)$

$(7, 1) (7, 2) (7, 3) (7, 4) (7, 5) (7, 6) (8, 1) (8, 2) (8, 3) (8, 4) (8, 5) (8, 6)$

$(9, 1) (9, 2) (9, 3) (9, 4) (9, 5) (9, 6) (10, 1) (10, 2) (10, 3) (10, 4) (10, 5) (10, 6)\}$

$n(S) = 60$ --- total outcomes

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

$x_1,x_2> 4 = \{(5, 5) (5, 6)(6, 5) (6, 6)(7, 5) (7, 6) (8, 5) (8, 6)(9, 5) (9, 6) (10, 5) (10, 6)\}$

$n(x_1,x_2>4) = 12$

So, the probability is:

$Pr = \frac{n(x_1,x_2>4)}{n(S)}$

$Pr = \frac{12}{60}$

$Pr = 0.20$