Question

A 10-sided die numbered 1 to 10 is rolled once. Find these probabilities.

Answer 1

Answer:

Remember that standard probabilities are defined as the ratio between the number of favorable cases, and the total number of possible events.

In this case, we have a 10-sided die, which means its faces are numbered from 1 to 10, which gives us 10 total number of possible events. In other words, the denominator of the ratio is going to be 10.

Now we're able to find each probability.

(A) Probability of getting an 8:

$P_{8} =\frac{1}{10}=0.10$

The numerator is 1 because there's only one number 8 in the die, that means the number of favorable cases is 1, and, as we said before, the total number of possible events is 10.

(B) Probability of getting an odd number:

$P_{odd} =\frac{5}{10}=0.5$

The numerator is 5 because there are 5 odd numbers from 1 to 10. In other words, there are 5 favorable cases to this probability.

(C) Probability of getting an even number:

$P_{even}=\frac{5}{10}=0.5$

There are 5 even numbers from 1 to 10, that's why we had the same probability.

(D) Probability of getting a number less than 6:

We know that there are only 5 numbers less than 6 for this die.

$P_{$

(E) Probability of getting a prime number:

A prime number is such that it can be divided only by itself and the unit. So, there are four prime numbers from 1 to 10, which are 2, 3, 5, 7.

$P_{prime}=\frac{4}{10}=0.4$

(F) Probability of getting 3 or 8:

When we are going to find the probability of an event "or" another, we must sum those favorable cases.

$P_{3 \ or \ 8} =\frac{1+1}{10}=\frac{2}{10}=0.2$

(G) Probability of getting 8, 9, or 10:

In this case, we need to some all three favorable cases. The die has only one 8, one 9 and one 10. So, the probability is

$P_{8,9,10}=\frac{1+1+1}{10}=\frac{3}{10} =0.3$

(H) Probability of getting a number greater than 9:

We know that there's only one number greater than 10 on such die.

$P_{>10}=\frac{1}{10}=0.1$

Lastly, "die" refers to this special 10-sided dice. In other problems, you can find "a die" with more faces even. But, in general, "die" refers to a dice.

Answer 2

Answer:

A. Pr(8)   = 1/10

B.  Pr(odd)  = 1/2

C.  Pr(even)  = 1/2

D. Pr(less than 6)  = 1/2

E.  Pr(prime) = 2/5

F. Pr(3 or 8)  = 1 / 5

G. Pr(8, 9 or 10) = 3/10

H. Pr(greater than 9)  = P(10) = 1/10

Step-by-step explanation:

We assume a FAIR 10-faced die, meaning there is an equal probability of throwing any one of the ten numbers.

With 10 possible outcomes each with equal probability, the probability of throwing any number is 1/10.

(It is preferable to work with fractions in probability because fractions are exact numbers, while decimals can often be rounded, or truncated).

A. Pr(8)   means probability of throwing an 8,  therefore Pr(8) = 1/10

B. Pr(odd)  there are 5 odd number from 1 to 10, so Pr(odd) = 5/10 = 1/2

C. Pr(even)  there are 5 even number from 1 to 10, so Pr(even) = 5/10 = 1/2

D. Pr(less than 6)  there are 5 numbers less than 6 (between1 to 10),

       so Pr(less than 6) = 5/10 = 1/2

E. Pr(prime) (remember that 1 is not prime)

   A prime number is an integer not divisible by any number except one and itself.

   Between 1 to 10, the four prime numbers are 2,3,5,7

   therefore Pr(prime) = 4/10 = 2/5

F. Pr(3 or 8)  

(3 or 8) make 2 successful outcomes out of 10, so Pr(3 or 8) = 2/10 = 1 / 5

G. Pr(8, 9 or 10)

similarly, (8,9 or 10) make three successful possible outcomes out of 10, so

P(8,9 or 10) = 3/10

H. Pr(greater than 9)

there is only one successful outcome, namely "10" out of 10 possible outcomes.  So

P(greater than 9) = P(10) = 1/10

1. please explain what’s a “die” all I know was “dice” but dice is 6 sided yea?

A die is singular for dice, which can take up any number of faces.  With a 10-faced solid, we can make a 10-faced die numbered 1 to 10.

2. and please explain the answer... I’m international student so I still need some explanation for the questions and answers thank you

See solutions above.