Answer:
What is the probability that he will win the game? 1/18
What is the probability that he will lose his next turn? 1/2
Step-by-step explanation:
All explanations here suppose the cubes are regular 6-facets cubes numbered 1 to 6. There are many different types of number cubes, and the question doesn't specify a type used.
What is the probability that he will win the game?
Out of the 36 possible combinations for 2 cubes to produce numbers, only 2 will give a 11. If the first cube gives a 6 and the second cube a 5, or vice-versa.
So, out of the 36 possible outcomes, only 2 are good... so 2/36 or 1/18.
What is the probability that he will lose his next turn?
For each time the first cube lands on 1, there are 3 possibilities that the number shown on the second cube (if that second cube shows 1, 3 or 5) will make an even total. That is true for each of the six possible values of the first cube.
So, in total, there are 18 possibilities out of 36 to throw an even sum, so 18/36 or 1/2.
Answer:
The probability that he will win the game is [tex]P_{11} = 0.0556[/tex]
The probability that he will lose his next turn is [tex]P_{even} = 0.5[/tex]
Step-by-step explanation:
By throwing a number cube you can get 6 different results. (1, 2, 3, 4, 5 or 6)
When throwing two numerical cubes, 6x6 = 36 possible results can be obtained.
The different ways to obtain a number 11 are:
[6] [5], [5] [6].
There are 2 possible ways out of 36.
Therefore the probability of winning the game is:
[tex]P_{11} = \frac{2}{36}\\\\P_{11} = 0.0556[/tex]
The different ways to obtain an even number are:
[1] [1], [2] [2], [3] [3], [4] [4], [5] [5], [6] [6], [1] [3], [ 3] [1], [1] [5], [5] [1], [2] [4],
[4] [2], [2] [6], [6] [2], [3] [5], [5] [3], [4] [6]. [6] [4]
There are 18 ways to get an even number.
Therefore the probability of losing the turn is:
[tex]P_{even} = \frac{18}{36}\\\\P_{even} = 0.5[/tex]
