WILL GIVE BRAINEST

What is the probability of flipping a coin 10 times and getting tails 7 times ? Round your answer to the nearest tenth of a perc

KatRina [158]

Answer:

Option C: 11.7%

Step-by-step explanation:

The number of elements in sample space when a coin is flipped 10 times will be:

$n(S) = 2^{10} \\= 1024$

Let A be the event that the tails appear 7 times:

Then,

$C_{(n,k)}=\frac{n!}{(k!)(n-k)!} \\where\\n=population\\k=picks\\In our case, k =7\\and\\n=10\\C_{(10,7)}= \frac{10!}{(7!)(10-3)!}\\= 120\\So, \\P(A) = \frac{n(A)}{n(S)} \\P(A) = \frac{120}{1024} \\= 0.1171\\$

Converting into percentage:

$P(A) = 0.1171*100=11.7%$

So, option C is correct ..

6 0

3 years ago

PLEASE HELP WITH THIS QUESTION, WILL MARK BRAINLIEST + 10 POINTS<br> #17

slega [8]

Sin 30 = 16/x
x = 16/sin 30 = 32
The answer is d: 32.

3 0

2 years ago

Plz answer. only if correct

earnstyle [38]

Answer:10001.5%+10303990=f(g+2)

Step-by-step explanation:

i dont remeber how to do this cap

5 0

3 years ago

Each Cube inside the rectangle Prisma has a edge length of 3/4 inches which is the value of the Prisma

Sphinxa [80]

Answer:

Step-by-step explanation:

6 0

3 years ago

student randomly receive 1 of 4 versions(A, B, C, D) of a math test. What is the probability that at least 3 of the 5 student te

alexdok [17]

Answer:

1.2%

Step-by-step explanation:

We are given that the students receive different versions of the math namely A, B, C and D.

So, the probability that a student receives version A = $\frac{1}{4}$ .

Thus, the probability that the student does not receive version A = $1-\frac{1}{4}$ = $\frac{3}{4}$ .

So, the possibilities that at-least 3 out of 5 students receive version A are,

1) 3 receives version A and 2 does not receive version A

2) 4 receives version A and 1 does not receive version A

3) All 5 students receive version A

Then the probability that at-least 3 out of 5 students receive version A is given by,

$\frac{1}{4}\times \frac{1}{4}\times \frac{1}{4}\times \frac{3}{4}\times \frac{3}{4}$ + $\frac{1}{4}\times \frac{1}{4}\times \frac{1}{4}\times \frac{1}{4}\times \frac{3}{4}$ + $\frac{1}{4}\times \frac{1}{4}\times \frac{1}{4}\times \frac{1}{4}\times \frac{1}{4}$