To win the jackpot, 4 different numbers are randomly selected from 1 to 46 and one number from 1 to 22. the order of the first 4 numbers does not matter. what is the probability of winning the jackpot on one try?

Probability of winning on one try : 1.16060875e-8

Step-by-step explanation:

For the first first 4 numbers.

Probability is 1/46 for the first number. Since the numbers are different, and it doesnt matter the order, the second number has a probability now of 1/45, the third has a probability of 1/44 and the last one a probability of 1/43.

Since the probability is dependan of the results of hitting the other number the probability of the first for numbers is the multiple of the 4 probabilities.

So it is 1/46 * 1/45 * 1/44 * 1/43 = 1 / 3916440

And that number is then multiplied by the probabilty of hitting the last number. 1/22

So the final probability is :

1/86161680   = 1.16060875e-8

The probability of winning the jackpot on one try is 2.78 * 10^-7

Step-by-step explanation:

There are 46 balls in total  (46-1)+1 = 46 . (b-a)+1 is the formula for number of elements between a and b included.

We need to find the number of combinations possible of 4 balls ( as order doesn't matter - 1234 is the sames as 2341) . So the number of possible combinations of 4 balls taken from 1 - 46  is given by

C= n!/(r!(n-r)!) where n is the number of possible balls = 46 and r the size of combination = 4 and ! is factorial ( ex 3! = 3*2*1) This gives for this case

C= n!/(r!(n-r)!) = 46!/(4!(46-4)!)= 163,185 combinations.

But as there is a fifth ball with (22-1)+1   = 22 posible options each combination must be multiplied by 22( for example 1234 22 is one but also 1234 10 is other)

163,185*22= 3,590,070  possibilities.

The probability of winning is 1 in  3,590,070  possibilities. or

p = 1/ 3,590,070= 2.78 * 10^-7

b) well because you get well

the answer is d. (1.2, -4.7) on edgn

hope this !