The probability is given by the number of prime numbers divided by the total amount numbers that can be formed.
A
a
12
B
b
5/6
C
c
1/3
Practice makes perfect
a
Consider the given digits.
1,2,3,5We are asked to find how many two-digit numbers can be formed by randomly selecting from the given digits without replacement. In this case, we are taking two digits from the given ones and the order we choose them in is important. Therefore, we can use the permutation formula to find how many numbers can be formed.
P(4,2)&=4!/(4-2)! &⇓ P(4,2)&=12
We can form 12 different numbers.
12,13,15,21,23,25,
31,32,35,51,52,53
b
In this part we are asked to find the probability that a two-digit number contains a 2 or a 5. Note that we can have a number containing 2 and 5 at the same time. Therefore, we have overlapping events. If A and B are overlapping events, P(A or B) is given as follows.
P(AorB)=& P(A)+P(B) [0.3em]
&-P(AandB)
This means that we first need to find P(2), P(5), and P(2and5). To do so, let's use the numbers we found in Part A.
12,13, 15, 21, 23, 25,
31, 32, 35, 51, 52, 53
Now we can find the probability of P(2), P(5), and P(2and5).
P(2)&= 6/12 [1em]
P(5)&= 6/12 [1em]
P(2and5)&= 2/12
With this values, we can now find P(2or 5).
P(2or 5)&= 6/12+ 6/12- 2/12
&⇓
P(2or 5)&=5/6
c
In this part we are asked to find the probability of getting a two-digit prime number. Let's first recall all of the different numbers we can form with the given digits.
12, 13,15,21, 23,25,
31,32,35,51,52, 53
Note that the total number of prime numbers is 4 and the number of different two-digit numbers that can be formed is 12. With this information we can now find the probability of P(prime).
P(prime)=4/12 ⇓ P(prime)=1/3
