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Alja [10]
3 years ago
6

A person tosses a fair coin until she obtains 2 heads in a row. She then tosses a fair die the same number of times as she tosse

d the coin. What is the expectation and variance for the number of 1s in the sequence of die rolls? How would your answer change if the coin was unfair, and the probability of obtaining heads was p?
Mathematics
1 answer:
ruslelena [56]3 years ago
5 0

Answer:

The expectation and variance for the number of 1s in the sequence of die rolls is:  1/6

If the coin was unfair the change is: 1/2

The probability of obtainig heads was: 51%

Step-by-step explanation:

In probability and statistics theory:

The expectation and variance possibilities gotten rolling one dice are six.

In independent Bernoulli trials sequence with probability 1/2 of success on each trial is metaphorically called a fair coin.

By the other hand if the probability is not 1/2, it is called a biased or unfair coin.

It might be thought  that the toss of a coin is always a 50/50 probability, with a 50 percent chance it lands on heads, and a 50 percent chance it lands on tails but if it starts out as heads, there's a 51% chance that it will end as heads.

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In a binomial distribution, n = 12 and π = .60.
Lemur [1.5K]

For binomial distribute we use formula

nCr (p)^r (q)^{n-r}

the probability mass function of the binomial distribution is:

nCr (\pi)^r (1-\pi)^{n-r}

Given n = 12 and π = .6. (Replace the values)

the probability for x = 5, we need to find P(x=5). Here r= 5

P(x=5)= 12C5(\pi )^5 (1-\pi)^{12-5}

= 12C5(0.6 )^5 (0.4)^{7}= 0.10090 = 0.101

(b) the probability for x ≤ 5

P(x<=5) = P(x=0) + P(x=1) +P(x=2)+ P(x=3) + P(x=4) + P(x=5)

P(x<=5) = 12C0(0.6 )^0(0.4)^{12} + 12C1(0.6 )^1(0.4)^{11} +  12C2(0.6 )^2(0.4)^{10} + 12C3(0.6 )^3(0.4)^{9} + 12C4(0.6 )^4(0.4)^{8} + 12C5(0.6 )^5 (0.4)^{7}

= 0.15821229 = 0.158

(c) find the probability for x ≥ 6

P(x>=6) =P(x=6) + P(x=7) +P(x=8)+ P(x=9) + P(x=10) + P(x=11)+ P(x=12)

P(x>=6) = 12C6(0.6 )^6(0.4)^{6} + 12C7(0.6 )^7(0.4)^{5} + 12C8(0.6 )^8(0.4)^{4} + 12C9(0.6 )^9(0.4)^{3} + 12C10(0.6 )^{10}(0.4)^{2} +   12C11(0.6 )^{11}(0.4)^{1}  +  12C12(0.6 )^{12}(0.4)^{0}

= 0.8417877 = 0.0842 (rounded to 3 decimal places)



3 0
3 years ago
If the area is 45 meters square, what is the minimum number of square-tiles with sides of 30 cm by 30 cm that are needed to cove
nexus9112 [7]
22500 total
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3 years ago
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Lostsunrise [7]
498 
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8 0
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Solve when c=7 and d=3 <br>5c-3d+11​
Studentka2010 [4]

Answer:

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6 0
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Y=x-7<br> 6x-6y=-7<br><br> elimination method
Archy [21]

Answer:

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Step-by-step explanation:

Let's write it in a more traditional form

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Same result you get by elimination. Multiply the first by 6 and either add or subtract - your choice the second. In either case you will get 0 = 35 or 0=49, which are obviously absurd

7 0
3 years ago
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