Mathematics: Pllllz i neeed helpp

Pllllz i neeed helpp - id9301975, decoreyjpaipxv

Ответ:

mikurrjurdan

25.03.2023

79.81% probability of winning.

Step-by-step explanation:

For each ticket, there are only two possible outcomes. Either you win, or you do not. The probability of winning in a ticket is independent of other tickets. So we use the binomial probability distribution to solve this question.

Binomial probability distribution

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

In which $C_{n,x}$ is the number of different combinations of x objects from a set of n elements, given by the following formula.

$C_{n,x} = \frac{n!}{x!(n-x)!}$

And p is the probability of X happening.

The odds of winning the prize is 1 : 53

This means that $p = \frac{1}{53}$

If I purchased 84 tickets what are the odds of winning in percentage.

To win, at least one of the tickets must have the prize, of 84 tickets, so n = 84 .

We have to find $P(X \geq 0)$ , which is given by:

$P(X \geq 0) = 1 - P(X = 0)$

In which

$P(X = 0) = C_{84,0}.(\frac{1}{53})^{0}.(\frac{52}{53})^{84} = 0.2019$

$P(X \geq 0) = 1 - P(X = 0) = 1 - 0.2019 = 0.7981$

79.81% probability of winning.

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