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Anit [1.1K]
3 years ago
12

A fair coin is tossed repeatedly. What is the probability that three heads in a row will occur before a tail followed by two hea

ds?
Mathematics
1 answer:
Masteriza [31]3 years ago
3 0

The probability that you can get N heads in a row would be:

Let <span>p</span> be the probability of flipping a heads. Let <span>x</span> be number of flips needed to achieve <span>h </span>consecutive heads. The solution is <span><span>E(x) = (<span><span>1−<span>p^h) / (</span></span><span><span>p^h</span>(1−p))</span></span></span></span>

This expression may be derived as follows. The probability of being successful immediately is <span><span>p^r.</span></span> However, one might get a tails immediately. In that case, the number of flips needed is <span><span>1+E(x) </span></span>(one flip has been used and we are back to the original position). We might get a heads and then a tails. In this case two flips have been used and we are back to the original position. Continue this up to <span><span>h−1</span></span> heads followed by a tails in which case <span>h</span> flips have been used and we are back to the original position.

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A sample size 25 is picked up at random from a population which is normally
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a) P(X < 99) = 0.2033.

b) P(98 < X < 100) = 0.4525

Step-by-step explanation:

To solve this question, we need to understand the normal probability distribution and the central limit theorem.

Normal Probability Distribution:

Problems of normal distributions can be solved using the z-score formula.

In a set with mean \mu and standard deviation \sigma, the z-score of a measure X is given by:

Z = \frac{X - \mu}{\sigma}

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the p-value, we get the probability that the value of the measure is greater than X.

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For a skewed variable, the Central Limit Theorem can also be applied, as long as n is at least 30.

Mean of 100 and variance of 36.

This means that \mu = 100, \sigma = \sqrt{36} = 6

Sample of 25:

This means that n = 25, s = \frac{6}{\sqrt{25}} = 1.2

(a) P(X<99)

This is the pvalue of Z when X = 99. So

Z = \frac{X - \mu}{\sigma}

By the Central Limit Theorem

Z = \frac{X - \mu}{s}

Z = \frac{99 - 100}{1.2}

Z = -0.83

Z = -0.83 has a pvalue of 0.2033. So

P(X < 99) = 0.2033.

b) P(98 < X < 100)

This is the pvalue of Z when X = 100 subtracted by the pvalue of Z when X = 98. So

X = 100

Z = \frac{X - \mu}{s}

Z = \frac{100 - 100}{1.2}

Z = 0

Z = 0 has a pvalue of 0.5

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Z = \frac{X - \mu}{s}

Z = \frac{98 - 100}{1.2}

Z = -1.67

Z = -1.67 has a pvalue of 0.0475

0.5 - 0.0475 = 0.4525

So

P(98 < X < 100) = 0.4525

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