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2 years ago

Who ever gets this right will get a brainlest and a good rating

Mathematics

2 answers:

Virty [35]2 years ago

5 0

It’s 60

I remember doing a test like that

Andrei [34K]2 years ago

4 0

Answer:

Step-by-step explanation:

Have a great day :)

( IF IT IS wrong sorry <3) but its prob correct i dunno

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Volume of a cube<br> 3.8 ft<br> 1.1 ft

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Step-by-step explanation:

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2 years ago

What is the value of StartFraction 1.6 times 10 Superscript 14 Baseline Over 3.2 times 10 Superscript 7 Baseline EndFraction in

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Answer:

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Step-by-step explanation:

i just took the test and i got it right.

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2 years ago

A coin, having probability p of landing heads, is continually flipped until at least one head and one tail have been flipped. (a

Natali [406]

Answer:

(a)

The probability that you stop at the fifth flip would be

$p^4 (1-p) + (1-p)^4 p$

(b)

The expected numbers of flips needed would be

$\sum\limits_{n=1}^{\infty} n p(1-p)^{n-1} = 1/p$

Therefore, suppose that $p = 0.5$ , then the expected number of flips needed would be 1/0.5 = 2.

Step-by-step explanation:

(a)

Case 1

Imagine that you throw your coin and you get only heads, then you would stop when you get the first tail. So the probability that you stop at the fifth flip would be

$p^4 (1-p)$

Case 2

Imagine that you throw your coin and you get only tails, then you would stop when you get the first head. So the probability that you stop at the fifth flip would be

$(1-p)^4p$

Therefore the probability that you stop at the fifth flip would be

$p^4 (1-p) + (1-p)^4 p$

(b)

The expected numbers of flips needed would be

$\sum\limits_{n=1}^{\infty} n p(1-p)^{n-1} = 1/p$

Therefore, suppose that $p = 0.5$ , then the expected number of flips needed would be 1/0.5 = 2.

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2 years ago