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eduard
3 years ago
10

You toss three 6-sided dice and record the sum of the three faces facing up. a) Describe precisely a sample space S for this exp

eriment. b) What is the probability that the sum is 16 or more? c) What is the probability the sum is exactly 4 or 5? d) What is the probability the sum is less than 17?
Mathematics
1 answer:
timama [110]3 years ago
3 0

Answer:

a.)Sample space means all possible outcomes, since we know the dice are 6 face, then the sample space becomes all possible outcomes when we toss the die.

b.) 9/216

c.) 9/216

d.) 212/216

Step-by-step explanation:

Sample space means all possible outcomes, since we know the dice are 6 face, then the sample space becomes all possible outcomes when we toss the die.

[1,1,1] [1,1,2] [1,1,3] [1,1,4] [1,1,5] [1,1,6]

[1,2,1] [1,2,2] [1,2,3] [1,2,4] [1,2,5] [1,2,6],

[1,3,1] [1,3,2] [1,3,3] [1,3,4] [1,3,5] [1,3,6]

[1,4,1] [1,4,2] [1,4,3] [1,4,4] [1,4,5] [1,4,6]

[1,5,1] [1,5,2] [1,5,3] [1,5,4] [1,5,5] [1,5,6]

[1,6,1] [1,6,2] [1,6,3] [1,6,4] [1,6,5] [1,6,6]

[2,1,1] [2,1,2] [2,1,3] [2,1,4] [2,1,5] [2,1,6]

[2,2,1] [2,2,2] [2,2,3] [2,2,4] [2,2,5] [2,2,6],

[2,3,1] [2,3,2] [2,3,3] [2,3,4] [2,3,5] [2,3,6]

[2,4,1] [2,4,2] [2,4,3] [2,4,4] [2,4,5] [2,4,6]

[2,5,1] [2,5,2] [2,5,3] [2,5,4] [2,5,5] [2,5,6]

[2,6,1] [2,6,2] [2,6,3] [2,6,4] [2,6,5] [2,6,6]

[3,1,1] [3,1,2] [3,1,3] [3,1,4] [3,1,5] [3,1,6]

[3,2,1] [3,2,2] [3,2,3] [3,2,4] [3,2,5] [3,2,6],

[3,3,1] [3,3,2] [3,3,3] [3,3,4] [3,3,5] [3,3,6]

[3,4,1] [3,4,2] [3,4,3] [3,4,4] [3,4,5] [3,4,6]

[3,5,1] [3,5,2] [3,5,3] [3,5,4] [3,5,5] [3,5,6]

[3,6,1] [3,6,2] [3,6,3] [3,6,4] [3,6,5] [3,6,6]

[4,1,1] [4,1,2] [4,1,3] [4,1,4] [4,1,5] [4,1,6]

[4,2,1] [4,2,2] [4,2,3] [4,2,4] [4,2,5] [4,2,6],

[4,3,1] [4,3,2] [4,3,3] [4,3,4] [4,3,5] [4,3,6]

[4,4,1] [4,4,2] [4,4,3] [4,4,4] [4,4,5] [4,4,6]

[4,5,1] [4,5,2] [4,5,3] [4,5,4] [4,5,5] [4,5,6]

[4,6,1] [4,6,2] [4,6,3] [4,6,4] [4,6,5] [4,6,6]

[5,1,1] [5,1,2] [5,1,3] [5,1,4] [5,1,5] [5,1,6]

[5,2,1] [5,2,2] [5,2,3] [5,2,4] [5,2,5] [5,2,6],

[5,3,1] [5,3,2] [5,3,3] [5,3,4] [5,3,5] [5,3,6]

[5,4,1] [5,4,2] [5,4,3] [5,4,4] [5,4,5] [5,4,6]

[5,5,1] [5,5,2] [5,5,3] [5,5,4] [5,5,5] [5,5,6]

[5,6,1] [5,6,2] [5,6,3] [5,6,4] [5,6,5] [5,6,6]

[6,1,1] [6,1,2] [6,1,3] [6,1,4] [6,1,5] [6,1,6]

[6,2,1] [6,2,2] [6,2,3] [6,2,4] [6,2,5] [6,2,6],

[6,3,1] [6,3,2] [6,3,3] [6,3,4] [6,3,5] [6,3,6]

[6,4,1] [6,4,2] [6,4,3] [6,4,4] [6,4,5] [6,4,6]

[6,5,1] [6,5,2] [6,5,3] [6,5,4] [6,5,5] [6,5,6]

[6,6,1] [6,6,2] [6,6,3] [6,6,4] [6,6,5] [6,6,6]

b.) Probability that the sum is 16 or more is

Pr[4,6,6] + pr[5,5,6] + pr [ 5,6,5] + pr [5,6,6] + pr [6,5,5] + pr [6,5,6] + pr [6,6,4] + pr[6,6,5] + pr [6,6,6]

Becomes:

[1/6]³ + [1/6]³ + [1/6]³ + [1/6]³ + [1/6]³ + [1/6]³ + [1/6]³ + [1/6]³ + [1/6]³ = 9/216

Probability that the sum is 4 or 5 is

Pr [ 1,1,2] or pr[1,2,2] or pr [1,1,3] or pr [1,2,1] or pr[2,1,2] or pr[1,3,1] or pr[3,1,1] or pr[2,1,1] or pr[2,2,1]

Becomes:

[1/6]³ + [1/6]³ + [1/6]³ + [1/6]³ + [1/6]³ + [1/6]³ + [1/6]³ + [1/6]³ + [1/6]³ = 9/216

Probability that the sum is less than 17

We take it as:

1- probability that the sum is 17 and above.

Now probability that the sum is 17 and above becomes

pr[5,6,6] or pr[6,5,6] or pr[6,6,5] or pr[6,6,6]

= [1/6]³ + [1/6]³ + [1/6]³ + [1/6]³ = 4/216

Hence, probability that the sum is less than 17 becomes:

1-4/216 = 212/216.

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