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elixir [45]
2 years ago
8

If 12 fair coins are flipped once, what is the probability of a result as extreme as or more extreme than 10 heads? (Note, "as o

r more extreme than" means in *either* direction -- that is to say, as or more extreme both in terms of getting that many or more heads, and in terms of getting that many or more tails.)
Mathematics
1 answer:
maw [93]2 years ago
3 0

Using the binomial probability concept ; the probability of obtaining 10 or more heads is 0.0457

Recall :

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

Where :

  • n = number of trials = 12
  • x ≥ 10
  • p = probability of success = 0.5
  • q = 1 - q = 0.5

P(x ≥ 10) = P(x = 10) + P(x = 11) + P(x = 12)

Using a binomial probability calculator :

P(x = 10) =

P(x = 11) =

P(x = 12) =

P(x ≥ 10) = 0.01611 + 0.02930 + 0.000244

P(x ≥ 10) = 0.045654

Therefore, probability of getting <em>as</em><em> </em><em>extreme</em><em> </em><em>or</em><em> </em><em>more</em><em> </em><em>extreme</em><em> </em><em>than</em><em> </em><em>10</em><em> </em><em>heads</em><em> </em><em>is</em><em> </em>0.0457.

Learn more :brainly.com/question/12474772

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3 years ago
Let C(n, k) = the number of k-membered subsets of an n-membered set. Find (a) C(6, k) for k = 0,1,2,...,6 (b) C(7, k) for k = 0,
vladimir1956 [14]

Answer:

(a) C(6,0) = 1, C(6,1) = 6, C(6,2) = 15, C(6,3) = 20, C(6,4) = 15, C(6,5) = 6, C(6,6) = 1.

(b) C(7,0) = 1, C(7,1) = 7, C(7,2) = 21, C(7,3) = 35, C(7,4) = 35, C(7,5) = 21, C(7,6) = 7, C(7,7)=1.

Step-by-step explanation:

In this exercise we only need to recall the formula for C(n,k):

C(n,k) = \frac{n!}{k!(n-k)!}

where the symbol n! is the factorial and means

n! = 1\cdot 2\cdot 3\cdot 4\cdtos (n-1)\cdot n.

By convention 0!=1. The most important property of the factorial is n!=(n-1)!\cdot n, for example 3!=1*2*3=6.

(a) The explanations to the solutions is just the calculations.

  • C(6,0) = \frac{6!}{0!(6-0)!} = \frac{6!}{6!} = 1
  • C(6,1) = \frac{6!}{1!(6-1)!} = \frac{6!}{5!} = \frac{5!\cdot 6}{5!} = 6
  • C(6,2) = \frac{6!}{2!(6-2)!} = \frac{6!}{2\cdot 4!} = \frac{5!\cdot 6}{2\cdot 4!} = \frac{4!\cdot 5\cdot 6}{2\cdot 4!} = \frac{5\cdot 6}{2} = 15
  • C(6,3) = \frac{6!}{3!(6-3)!} = \frac{6!}{3!\cdot 3!} = \frac{5!\cdot 6}{6\cdot 6} = \frac{5!}{6} = \frac{120}{6} = 20
  • C(6,4) = \frac{6!}{4!(6-4)!} = \frac{6!}{4!\cdot 2!} = frac{5!\cdot 6}{2\cdot 4!} = \frac{4!\cdot 5\cdot 6}{2\cdot 4!} = \frac{5\cdot 6}{2} = 15
  • C(6,5) = \frac{6!}{5!(6-5)!} = \frac{6!}{5!} = \frac{5!\cdot 6}{5!} = 6
  • C(6,6) = \frac{6!}{6!(6-6)!} = \frac{6!}{6!} = 1.

(b) The explanations to the solutions is just the calculations.

  • C(7,0) = \frac{7!}{0!(7-0)!} = \frac{7!}{7!} = 1
  • C(7,1) = \frac{7!}{1!(7-1)!} = \frac{7!}{6!} = \frac{6!\cdot 7}{6!} = 7
  • C(7,2) = \frac{7!}{2!(7-2)!} = \frac{7!}{2\cdot 5!} = \frac{6!\cdot 7}{2\cdot 5!} = \frac{5!\cdot 6\cdot 7}{2\cdot 5!} = \frac{6\cdot 7}{2} = 21
  • C(7,3) = \frac{7!}{3!(7-3)!} = \frac{7!}{3!\cdot 4!} = \frac{6!\cdot 7}{6\cdot 4!} = \frac{5!\cdot 6\cdot 7}{6\cdot 4!} = \frac{120\cdot 7}{24} = 35
  • C(7,4) = \frac{7!}{4!(7-4)!} = \frac{6!\cdot 7}{4!\cdot 3!} = frac{5!\cdot 6\cdot 7}{4!\cdot 6} = \frac{120\cdot 7}{24} = 35
  • C(7,5) = \frac{7!}{5!(7-2)!} = \frac{7!}{5!\cdot 2!} = 21
  • C(7,6) = \frac{7!}{6!(7-6)!} = \frac{7!}{6!} = \frac{6!\cdot 7}{6!} = 7
  • C(7,7) = \frac{7!}{7!(7-7)!} = \frac{7!}{7!} = 1

For all the calculations just recall that 4! =24 and 5!=120.

6 0
2 years ago
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svetlana [45]

Answer:

The unit rate is 0. 14 miles per minute.

Hence, option a is correct.

Step-by-step explanation:

Given

Running of 5/4 a mile in 9 minutes.

To determine

The unit rate = ?

Given that the running 5/4 a mile in 9 minutes.

i.e.

Miles covered in 9 minutes = 5/4 or 1.25 miles

Miles covered in 1 minute = 1.25/9

                                           = 0. 14 miles per minute

Therefore, we conclude that:

The unit rate is 0. 14 miles per minute.

Hence, option a is correct.

6 0
2 years ago
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