Suppose that a balanced, six-sided die has been altered so that the faces are 1, 2, 3, 4, 5, and 5. If the die is tossed five times, and the number on the uppermost face is recorded each time, what is the probability that the numbers recorded are 1, 2, 3, 4, and 5 in any order

0.0309

Step-by-step explanation:

Faces on dice ( 1, 2, 3, 4, 5, 5)

Number of outcomes = 6

With 5 repeated twice

P(event A) = Required outcome / Total possible outcomes

Required outcome :

Number of 5's = 2 * distinct faces!

Required outcome = 2 * 5! = 2 * 120 = 240

Total possible outcomes = (number of faces)^distinct Numbers = 6^5 = 7776

Required / total possible = 240 / 7776

= 0.0308641

= 0.0309

3.76%

step-by-step explanation:

i do not know this sorry : (