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3 years ago

If you flip a fair coin 7 times, what is the probability that you will get exactly 4 tails?

Mathematics

1 answer:

andre [41]3 years ago

7 0

Probability that you will get exactly 4 tails, if you flip a fair coin 7 times is: $\frac{35}{128}$

Step-by-step explanation:

We need to find the probability that you will get exactly 4 tails, if you flip a fair coin 7 times.

This is the combination question.

The total combinations will be: 2^7=128

Now, finding number of ways we can get exactly 4 tiles:

We will use the formula:

$^nC_k=\frac{n!}{k!(n-k)!}$

in our given question, n= 7, k= 4

Putting values:

$^nC_k=\frac{n!}{k!(n-k)!}\\^7C_4=\frac{7!}{4!*3!}\\ ^7C_4=\frac{7*6*5*4!}{4!*3!}\\ ^7C_4=\frac{7*6*5}{6}\\ ^7C_4=35$

So, Probability that you will get exactly 4 tails, if you flip a fair coin 7 times is: $\frac{35}{128}$

Keywords: Probability

Learn more about Probability at:

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A 90% confidence interval for the mean BAC in fatal crashes in which the driver had a positive BAC is [0.143, 0.177] .

Step-by-step explanation:

We are given that a researcher randomly selects records from 60 such drivers in 2009 and determines the sample mean BAC to be 0.16 g/dL with a standard deviation of 0.080 g/dL.

Firstly, the pivotal quantity for finding the confidence interval for the population mean is given by;

P.Q. = $\frac{\bar X-\mu}{\frac{s}{\sqrt{n} } }$ ~ $t_n_-_1$

where, $\bar X$ = sample mean BAC = 0.16 g/dL

s = sample standard deviation = 0.080 g/dL

n = sample of drivers = 60

$\mu$ = population mean BAC in fatal crashes

Here for constructing a 90% confidence interval we have used a One-sample t-test statistics because we don't know about population standard deviation.

So, a 90% confidence interval for the population mean, $\mu$ is;

P(-1.672 < $t_5_9$ < 1.672) = 0.90 {As the critical value of t at 59 degrees of

freedom are -1.672 & 1.672 with P = 5%} P(-1.672 < $\frac{\bar X-\mu}{\frac{s}{\sqrt{n} } }$ < 1.672) = 0.90

P( $-1.672 \times {\frac{s}{\sqrt{n} } }$ < ${\bar X-\mu}$ < $1.672 \times {\frac{s}{\sqrt{n} } }$ ) = 0.90

P( $\bar X-1.672 \times {\frac{s}{\sqrt{n} } }$ < $\mu$ < $\bar X+1.672 \times {\frac{s}{\sqrt{n} } }$ ) = 0.90

90% confidence interval for $\mu$ = [ $\bar X-1.672 \times {\frac{s}{\sqrt{n} } }$ , $\bar X+1.672 \times {\frac{s}{\sqrt{n} } }$ ]

= [ $0.16-1.672 \times {\frac{0.08}{\sqrt{60} } }$ , $0.16+1.672 \times {\frac{0.08}{\sqrt{60} } }$ ]

= [0.143, 0.177]

Therefore, a 90% confidence interval for the mean BAC in fatal crashes in which the driver had a positive BAC is [0.143, 0.177] .

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3 years ago