The chance of winning a certain game at a carnival is 2 in 5. If Andy plays the game 12 times, what is the probability that he l

Question

Lunna

16 days ago

9

The chance of winning a certain game at a carnival is 2 in 5. If Andy plays the game 12 times, what is the probability that he l

oses AT MOST 3 times?

Mathematics

1 answer:

PIT_PIT [9K] · Answer 1 · 2026-03-11T23:25:14+03:00

Answer:

There is a 1.5267% chance that he loses AT MOST 3 times.

Step-by-step explanation:

Each game Andy plays has only two outcomes: a win or a loss. The probability of winning doesn’t depend on previous games, leading us to use the binomial probability distribution for this situation.

Binomial probability distribution

This distribution calculates the probability of achieving exactly x successes in n trials with only two possible outcomes for X.

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

Where $C_{n,x}$ signifies the various combinations of x objects from a set of n elements, calculated using:

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

And p represents the likelihood of X occurring.

The winning probability for a certain carnival game is 2 out of 5.

This implies a losing chance of (5-2) in 5, equaling 3 out of 5.

Thus, $p = \frac{3}{5} = 0.6$

Considering 12 games:

Consequently, $n = 12$ .

What probability is there that he loses AT MOST 3 times?

$P(X \leq 3) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3)$ .

Where:

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

$P(X = 0) = C_{12,0}.(0.6)^{0}.(0.4)^{12} = 0.000017$

$P(X = 1) = C_{12,1}.(0.6)^{1}.(0.4)^{11} = 0.000302$

$P(X = 2) = C_{12,2}.(0.6)^{2}.(0.4)^{10} = 0.002491$

$P(X = 3) = C_{12,3}.(0.6)^{3}.(0.4)^{9} = 0.012457$

$P(X \leq 3) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) = 0.000017 + 0.000302 + 0.002491 + 0.012457 = 0.015267$

The likelihood of losing AT MOST 3 times is 1.5267%.