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umka21 [38]
3 years ago
15

Suppose we have an unfair coin that its head is twice as likely to occur as its tail. a)If the coin is flipped 3 times, what is

the probability of getting exactly 1 head? Give your answer to four decimal places b)If the coin is flipped 5 times, what is the probability of getting exactly 2 tails? Give your answer to four decimal places c)If the coin is flipped 4 times, what is the probability of getting at least 3 tails? Give your answer to four decimal places
Mathematics
1 answer:
Alenkinab [10]3 years ago
8 0

Answer: a) 0.2222, b) 0.3292, c) 0.1111

Step-by-step explanation:

Since we have given that

Let the probability of getting head be p.

Since,  its head is twice as likely to occur as its tail.

p+\dfrac{p}{2}=1\\\\\dfrac{3p}{2}=1\\\\p=\dfrac{2}{3}

a)If the coin is flipped 3 times, what is the probability of getting exactly 1 head?

So, here, n  = 3

p=\dfrac{2}{3}

q=\dfrac{1}{3}

Now,

P(X=1)=^3C_1(\dfrac{2}{3})^1(\dfrac{1}{3})^2=0.2222

b)If the coin is flipped 5 times, what is the probability of getting exactly 2 tails?

2 tails means 3 heads.

So, it becomes,

P(X=3)=^5C_3(\dfrac{2}{3})^3(\dfrac{1}{3})^2=0.3292

c)If the coin is flipped 4 times, what is the probability of getting at least 3 tails?

P(X\leq 1)=\sum _{x=0}^1^4C_x(\dfrac{2}{3}^x(\dfrac{1}{3})^{4-x}=0.1111

Hence, a) 0.2222, b) 0.3292, c) 0.1111

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Answer:

See explanation

Step-by-step explanation:

Let x be the number of spade shovels, y -the number of flat shovels and z - the number of square showels sold that day.

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a) The system of three equations is

\left\{\begin{array}{l}x+y+z=80\\ \\x=2z\\ \\16x+9.60y+12.80z=1,072\end{array}\right.

b) In matrix form this is

\left(\begin{array}{ccc}1&1&1\\ 1&0&-2\\ 16&9.60&12.80\end{array}\right)\cdot \left(\begin{array}{c}x\\y\\z\end{array}\right)=\left(\begin{array}{c}80\\0\\1,072\end{array}\right)

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\left\|\begin{array}{ccc}1&1&1\\ 1&0&-2\\ 16&9.60&12.80\end{array}\right\|=0-32+9.60-0+19.20-12.80=-16

d) Find three determinants:

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\left\|\begin{array}{ccc}1&80&1\\ 1&0&-2\\ 16&1,072&12.80\end{array}\right\|=0-2,560+1,072-0+2,144-1,024=-368

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x=\dfrac{-608}{-16}=38\\ \\y=\dfrac{-368}{-16}=23\\ \\z=\dfrac{-304}{-16}=19

e) If the store doubled all prices and inventory, then the new matrix is

\left(\begin{array}{ccc}1&1&1\\ 1&0&-2\\ 32&19.20&25.60\end{array}\right)\cdot \left(\begin{array}{c}x\\y\\z\end{array}\right)=\left(\begin{array}{c}160\\0\\2,144\end{array}\right)

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