Question

A representative from a company that manufactures items for left-handed people will attend a large convention. The representative hopes to find a left-handed person at the convention to try out the items. The representative will select an attendee at random until a left-handed person is found. Assume each selection is independent of another. If 10 percent of the convention attendees are left-handed, what is the probability that the representative must select 4 attendees to find one who is left-handed

Answer 1

Answer: .1(0.9^3)

Step-by-step explanation:

We are given the following information:

We treat left banded as a success.

P(left handed) = 10% = 0.1

Then the number of attendees follows a geometric distribution, where

P(X=x) = p(1-p)^x-1

where x is the number of trails to get first success.

Now, we are given x = 4

=.1(.9)^3

Answer 2

The probability that the representative must select 4 attendees till a left handed person ls selected using the Geometric distribution principle is 0.0829

Given the following :

• Probability of being left handed, LH = p = 10% = 0.1

• 1 - p = 1 - 0.1 = 0.9

• Number of selections , x = 4

Using the Geometric distribution principle :

$probability \: = p {(1 - p)}^{x - 1}$

P(selection made = 4) = 0.1(0.9)³

P(selection made = 4) = 0.1 × 0.729 = 0.0729

Therefore, the probability that 4 selections is made is 0.0729

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