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Statistics and Probability

hira farrukh

Answered question

2022-07-18

suppose we go door to door selling candy. we consider it a success if someone buys a candy bar the probability that any given person will buy the candy bar is 0.4. What is the probability of experiencing 8 failures before a total of 5 successes.

Answer & Explanation

Andre BalkonE

Skilled2023-05-23Added 110 answers

To solve the problem, we can use the concept of a negative binomial distribution. The negative binomial distribution represents the number of failures before a specified number of successes occur.
In this case, we want to find the probability of experiencing 8 failures before a total of 5 successes. The probability of success, denoted by

p

, is given as 0.4.
Let's denote the random variable representing the number of failures before 5 successes as

X

. We want to find

P (X = 8)

.
The probability mass function (PMF) of the negative binomial distribution is given by:

P (X = k) = (\binom{k + r - 1}{k}) \cdot p^{r} \cdot (1 - p)^{k}

where

k

is the number of failures,

r

is the number of successes,

p

is the probability of success, and

(1 - p)

is the probability of failure.
In our case,

k = 8

r = 5

, and

p = 0.4

. Substituting these values into the PMF formula, we have:

P (X = 8) = (\binom{8 + 5 - 1}{8}) \cdot {0.4}^{5} \cdot (1 - 0.4)^{8}

Calculating the values within the binomial coefficient and simplifying the expression, we get:

P (X = 8) = (\binom{12}{8}) \cdot {0.4}^{5} \cdot {0.6}^{8}

Using the binomial coefficient formula

(\binom{n}{k}) = \frac{n!}{k! (n - k)!}

, we can calculate the binomial coefficient as follows:

(\binom{12}{8}) = \frac{12!}{8! (12 - 8)!} = \frac{12!}{8! 4!} = \frac{12 \cdot 11 \cdot 10 \cdot 9}{4 \cdot 3 \cdot 2 \cdot 1} = 495

Now, substituting this value back into the expression, we have:

P (X = 8) = 495 \cdot {0.4}^{5} \cdot {0.6}^{8}

Calculating the numerical value, we find:

P (X = 8) \approx 0.0060

Therefore, the probability of experiencing 8 failures before a total of 5 successes is approximately 0.0060.

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