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 loses AT MOST 3 times?

1.5267% probability that he loses AT MOST 3 times

Step-by-step explanation:

For each game that Andy plays, there are only two possible outcomes. Either he wins, or he loses. The probability of winning a game is independent of other games. 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.

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

And p is the probability of X happening.

The chance of winning a certain game at a carnival is 2 in 5.

So the chance of losing is (5-2) in 5, that is 3 in 5.

So

12 games:

This means that .

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

.

In which:

1.5267% probability that he loses AT MOST 3 times

an

y = 6x -3y -12

y +3y +12 = 6x

=> . x = { 2y / 3 +2 } this is equation of line in terms of x

=> . if line meets the x-axis the co-ordinates of meeting point would be ( x , 0 )

=> . x = 2 ( 0 ) / 3 + 2

=> . x = 2

so, the point is ( 2 , 0 )

answer: b

step-by-step explanation: im sorry bro bro but thats not a eqation the human mind can answer and plus im a devil