Suppose the chance of winning a game is 1/50. if you play the game 50 times independently, is it certain that you will win? if not, what is the chance of winning at least once? a.0.0200 b.0.9462 c.0.6358 d.1.0000

We are given chance of winning a game is 1/50.

In decimal form 1/50 could be written as 0.02.

So, for each game, the chance of winning is 0.02.

Therefore, chance of loss would be = 1 -(0.02) = 0.98.

Chance of loss of 50 independent games = 0.98*0.98*0.98*0.98 upto 50 times, that is (0.98)^50.

The chance of winning at least once = 1 - (0.98)^50.

We got (0.98)^50 = 0.36417

Therefore, 1 - (0.98)^50 = 1 - 0.36417.

We get 0.63583.

Therefore, correct option is c. 0.6358.

2.   2 3/5.

3.   2 1/2.

step-by-step explanation:

6 - 3 2/5

= 6 -   17/5

= 30/5 - 17/5

= 13/5

= 2 3/5 (answer).

5 1/4 - 2 3/4

= 21/4 - 11/4

= 10/4

= 2 1/2 (answer).

n=9

step-by-step explanation:

7 times a number = 7n

decreased by 13 = 7n-13

same as is =

5 times the number = 5n

increased by 5 = 5n+5

7n-13 = 5n+5

2n=18

n=9