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Andru [333]
3 years ago
5

A binomial distribution

Mathematics
1 answer:
Alina [70]3 years ago
3 0

Answer:

0.324

Step-by-step explanation:

Given that :

Success rate = 30%

p = 30% = 0.3

q = 1 - p = 1 - 0.3 = 0.7

Number of trials, n = 6

Probability of having exactly 2 successes ; x = 2

P(x = 2)

Usibgbtge binomial probability relation :

P(x = x) = nCx * p^x * q^(n-x)

P(x = 2) = 6C2 * 0.3^2 * 0.7^4

P(x = 2) = 15 * 0.3^2 * 0.7^4

P(x = 2). = 0.324135

P(x = 2) = 0.324

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Pete, a professional bowler, is unhappy with any game below 200. Over time, 80% of his games exceed this score. What is the prob
Aneli [31]

The probability that Pete exceeds 200 in at least 9 of his next 10 games is 0.268+0.1074 = 0.3754

<h3>What is Probability ?</h3>

Probability is defined as the likeliness of an event to happen.

It has a range of 0 to 1.

It is given that

Pete, a professional bowler, is unhappy with any game below 200

80% of his games exceed this score.

the probability that Pete exceeds 200 in at least 9 of his next 10 games is given by

P( 9/10 > 200) +P(10/10>200)

The binomial experiment consists of n trial out of it x is success.

P( x) = \dfrac{n!}{x!(n-x)!}p^x(1-p)^{n-x}

Here p = 0.8

For 9/10 matches

P( 9/10 > 200) = \dfrac{10!}{9!(10-9)!}0.8^9(1-0.8)^{10-9}\\\\ = {10}(0.8)^9(0.2)\\= 0.268

For (10/10 >200)

P( 10/10 > 200) = \dfrac{10!}{10!(10-10)!}0.8^{10} (1-0.8)^{10-10}\\\\ = {1!}(0.8)^{10}\\= 0.1074

Therefore the probability that Pete exceeds 200 in at least 9 of his next 10 games is 0.268+0.1074 = 0.3754.

To know more about Probability

brainly.com/question/11234923

#SPJ1

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