0
$\begingroup$

If I enter a contest with a 5% chance of winning. If I enter 4 more contests,(totaling 5 contests) each with the same 5% chance of winning. How do I calculate the chance of winning ONE of them?

  • 0
    *exactly* one, or *at least* one – 2017-02-01
  • 0
    So, have you studied **Binomial Distributions** yet? – 2017-02-01

1 Answers 1

0

Your chance of winning each contest = $\frac {5}{100} = \frac 1{20}$

Chance of not winning = $\frac {95}{100} = \frac {19}{20}$

If exactly one contest you win -

$\frac 1{20} × \frac {19}{20} × \frac {19}{20} × \frac {19}{20} × \frac {19}{20}$

Now multiply it with 5 as its not necessary you win first contest. It can be any 1 out of 5.

  • 0
    So I did .05 x .95 x .95 x .95 x .95? And got .0407 and then I multiply that by 5 for the total number of contests? And get .20 so the answer is 20%? – 2017-02-01
  • 0
    Yes .2036 and appr 20℅. – 2017-02-01
  • 0
    Ok and why do I multiply the chances of not winning? What formula is this following? – 2017-02-01
  • 0
    Its like you win first contest and loose other four contests. So we multiply probabilities. – 2017-02-01