If eli is guessing when he chooses a super bowl winner (that is, he mentally flips a fair coin to decide which team he chooses as the winner), what is the probability that eli chooses 7 winners in a row?

Respuesta :

It's even because the coin has 2 different sides, none of the sides make it more likely or less likely to choose 7 winners in a row.

The probability that Eli chooses [tex]7[/tex] winners in a row is [tex]\boxed{\bf 0.0078}[/tex].

Further explanation:

Given:

Eli flips a fair coin to decide which team chooses as a winner.

Concept used:

The probability [tex]P(A)[/tex] of any event [tex]A[/tex] can be calculated as follows:

[tex]\boxed{P(A)=\dfrac{n(A)}{n(A)}}[/tex] …… (1)

Here, [tex]n(A)[/tex] is the total number of elements in event [tex]A[/tex] and [tex]n(S)[/tex] is the number of element in sample space of an experiment.

Calculation:

The sample space is the total possible outcomes in an experiment.

Consider [tex]n(S)[/tex] as the number of element in sample space [tex]S[/tex].

The possible outcomes in sample space [tex]S[/tex] are win and lose.

The sample space [tex]S[/tex] is as follows:

[tex]\boxed{S=\{H,T\}}[/tex]

Therefore, the number of element in sample space [tex]S[/tex] is as follows:

[tex]\boxed{n(S)=2}[/tex]  

Consider [tex]A[/tex] as the event that Eli choosing the winner team, [tex]n(A)[/tex] as the number of favorable outcomes in event [tex]A[/tex] and  [tex]P(A)[/tex] as the probability of event [tex]A[/tex].

The winner is chooses if Eli flips the coin. There is chance that Eli chooses winner either head comes or tail comes.

Therefore, there is only one favorable outcomes to choose the winning team.

The number of favorable outcomes in an event [tex]A[/tex] can be calculated as follows:

[tex]\boxed{n(A)=1}[/tex]  

Substitute [tex]1[/tex] for [tex]n(A)[/tex] and [tex]2[/tex] for [tex]n(S)[/tex] in the equation (1) to obtain the probability [tex]P(A)[/tex].

[tex]\boxed{P(A)=\dfrac{1}{2}}[/tex]  

The probability that Eli chooses [tex]7[/tex] winner in a row can be calculated as follows:

[tex]\begin{aligned}P(A)&=\left(\dfrac{1}{2}\cdot \dfrac{1}{2}\cdot \dfrac{1}{2}\cdot \dfrac{1}{2}\cdot \dfrac{1}{2}\cdot \dfrac{1}{2}\cdot \dfrac{1}{2}\right)\\&=\dfrac{1}{2^{7}}\\&=\dfrac{1}{128}\\&=0.0078\end{aligned}[/tex]

 

Thus, the probability that Eli chooses [tex]7[/tex] winner in a row is [tex]\boxed{\bf 0.0078}[/tex].

Learn more:

1. Learn more about problem on circle https://brainly.com/question/9510228

2. Learn more about problem on graph https://brainly.com/question/2334270

Answer details:

Grade: Senior school

Subject: Mathematics

Chapter: Probability

Keywords: Probability, Eli, favorable outcomes, event, experiment, winner, sample space, fair coin, number of element, 7 winners, in a row, basic probability theorem.