+0

# NT problem

0
81
5
+83

What is the probability that the same number will be facing up on each of three standard six-sided dice that are tossed simultaneously? Express your answer as a common fraction.

Jul 25, 2020

#1
0

First die =any number =6/6

2nd die - The probability that it is different from the first die = 5/6

3rd die -  The probability that it is different from the first two= 4/6

Then, the probability that ALL three are different from each other =6/6  x  5/6  x  4/6 =120 / 216

Then, the probability that ALL three show the same number  = 1 - 120 / 216 = 96 / 216 = 4 / 9

Jul 25, 2020
#2
+26703
+1

216 possible rolls

6 rolls will fit the criteria   666 555 444 333 222 111

6 /216 = 1/36 chance

Jul 25, 2020
edited by ElectricPavlov  Jul 25, 2020
#3
0

(1)  The first die can be anything.

(2)  The odds of the second die being the same as the first  =  1/6

(3)  The odds of the third      die being the same as the first  =  1/6

1        1          1

The odds of both (2) and (3) happening  =   ––  x  ––   =   —–

6       6          36

.

Jul 25, 2020
#4
+60
0

There is a 1/35 chance

Jul 25, 2020
#5
0

What is this, "A for Effort" day?  The answer is wrong and somebody gave it a "plus heart"?

Guest Jul 26, 2020
{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}