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3 years ago

15

A pair of dice are rolled. What is the probability of rolling 4 or greater?

1 answer:

skelet666 [1.2K]3 years ago

6 0

Answer: option D: 11/12

Explanation:

1) The sample space is formed by 36 pairs: 6×6 = 36.

2) Roolling 4 or more is the same that not rolling 3 or less

3) Rolling 3 or less are (1,1), (1,2), and (2,1). Those are 3 outcomes out of 36.

4) Therefore, not rolling 3 or less (the same that rolling 4 or more) has 36 - 3 = 33 outcomes out of 36.

5) The probability of 33 out of 36 is 33/36

6) Simplify the fraction 33/36 = 11/12

And that is the answer.

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3 years ago

What is the probability of rolling a four then a three

spayn [35]

Answer:

$\frac{1}{36}$

Step-by-step explanation:

Total possibilities when we a roll a die at a time are 6

given we should have four for first time and then three

let us assume we rolled the dice we may get 1,2,3,4,5,6(any of these) the probability to get 4 is

PROBABILITY= $\frac{\textrm{FAVOURABLE CHANCES}}{\textrm{TOTAL CHANCES}}$

Favourable chances=1

Total chances=6

Probability= $\frac{1}{6}$

the prabability to get 4 in first roll is $\frac{1}{6}$ .

let us assume we rolled the dice for second time again we may get 1,2,3,4,5,6(any of these) the probability to get 3 is

Favourable chances=1

total chances=6

probability= $\frac{1}{6}$

the probability to get 3 in second roll irrespective of first one is $\frac{1}{6}$

the probability to get 4 in first time and then 3 is

The probability to occur both events at a time is multiplication of individual probabilities

So,

probablility to get 4 in first roll= $\frac{1}{6}$

probability to get 3 in second roll= $\frac{1}{6}$

probability to occur both at a same time is = $\frac{1}{6}$ $\times$ $\frac{1}{6}$ = $\frac{1}{36}$

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2 years ago