Find the probability that when four numbers from 1 to 100, inclusive, are picked at random with no repetitions allowed, either all are odd, all are divisible by 3, or all are divisible by 5.

The probability is;

0.076

Step-by-step explanation:

The count of numbers from 1 to 100 is 100

Let’s find the number of odd numbers

We can use;

a + (n-1)d = T

a is 1, d is 2 and T is 99

Thus;

1 + 2(n-1) = 99

n-1 = (99-1)/2

n-1 = 59

n = 50

So there are 50 odd numbers

Probability of selecting an odd number is 50/100 = 1/2

If all 4 are odd, we have 1/2^4 = 1/16

Let’s get the multiples of 3

we can use

a + (n-1)d = T

a = 3, n = ? , d = 3 and T = 99

So;

3 + 3(n-1) = 99

n-1 = 34

n = 33

So the probability is 33/100

All 4 are divisible will give 33/100^4

Multiples of 5

a + (n-1)d = T

a = 5, n = ? , d = 5 T = 100

5 + (n-1)5 = 100

5(n-1) = 95

n-1 = 19

n = 20

Probability is 20/100 = 1/5

All

four are multiples of 5

1/5^4 = 1/625

So the probability is;

1/16 + 1/625 + 0.01185921

= 0.0625 + 0.0016 + 0.01185921 = 0.07595921

y=3n + 1

step-by-step explanation:

this is wrong because it says the second number is one less and i added one instead