Question

Before 1918, approximately 60% of the wolves in a region were male, and 40% were female. However, cattle ranchers in this area have made a determined effort to exterminate wolves. From 1918 to the present, approximately 70% of wolves in the region are male, and 30% are female. Biologists suspect that male wolves are more likely than females to return to an area where the population has been greatly reduced. (Round your answers to three decimal places.) (a) Before 1918, in a random sample of 12 wolves spotted in the region, what is the probability that 9 or more were male? What is the probability that 9 or more were female? What is the probability that fewer than 6 were female? (b) For the period from 1918 to the present, in a random sample of 12 wolves spotted in the region, what is the probability that 9 or more were male? What is the probability that 9 or more were female? What is the probability that fewer than 6 were female?

Answer 1

Answer:

A) 0.2253, 0.0153; B) 0.4925, 0.0017

Step-by-step explanation:

This is a binomial distribution.  This is because there are only two outcomes; each trial is independent of each other; and the outcomes are independent.

This means we use the formula

$_nC_r\times p^r\times (1-p)^{n-r}$

For part A,

There are 12 wolves selected; this means n = 12.  We want the probability that 9 or more are male; this makes r = 9, 10, 11 or 12.  We will find each probability and add them together.

p, the probability of success, is 0.6 for the first question (males).  This makes 1-p = 1-0.6 = 0.4.  Together this gives us

$_{12}C_9(0.6)^9(0.4)^3+_{12}C_{10}(0.6)^{10}(0.4)^2+_{12}C_{11}(0.6)^{11}(0.4)^1+_{12}C_{12}(0.6)^{12}(0.4)^0\\\\=220(0.6)^9(0.4)^3+66(0.6)^{10}(0.4)^2+12(0.6)^{11}(0.4)+1(0.6)^{12}(1)\\\\\\= 0.2253$

We now want the probability that 9 or more are female; this makes r = 9, 10, 11 or 12.  p is now 0.4; this makes 1-p = 1-0.4 = 0.6.  This gives us

$_{12}C_9(0.4)^9(0.6)^3+_{12}C_{10}(0.4)^{10}(0.6)^2+_{12}C_{11}(0.4)^{11}(0.6)^1+_{12}C_{12}(0.4)^{12}(0.6)^0\\\\=220(0.4)^9(0.6)^3+66(0.4)^{10}(0.6)^2+12(0.4)^{11}(0.6)^1+1(0.4)^{12}(1)\\\\=0.0153$

For part B,

There are again 12 wolves selected, so n = 12.  We want the probability in the first question that 9 or more are male; this makes r = 9, 10, 11 or 12.  The probability of success is now 0.7, so 1-p = 1-0.7 = 0.3$_{12}C_9(0.7)^9(0.3)^3+_{12}C_{10}(0.7)^{10}(0.3)^2+_{12}C_{11}(0.7)^{11}(0.3)^1+_{12}C_{12}(0.7)^{12}(0.3)^0\\\\=220(0.7)^9(0.3)^3+66(0.7)^{10}(0.3)^2+12(0.7)^{11}(0.3)^1+1(0.7)^{12}(0.3)^0\\\\= 0.4925$

For the second question, the probability of success is now 0.3 and 1-p = 1-0.3 = 0.7:

$220(0.3)^9(0.7)^3+66(0.3)^{10}(0.7)^2+12(0.3)^{11}(0.7)^1+1(0.3)^{12}(0.7)^0\\\\=0.0017$

Answer 2

Probabilities are used to determine the outcomes of events.

Before 1918,

• The probability of selecting 9 or more male wolves is 0.225
• The probability of selecting 9 or more female wolves is 0.015
• The probability of selecting lesser than 6 female wolves is 0.665

Since 1918,

• The probability of selecting 9 or more male wolves is 0.493
• The probability of selecting 9 or more female wolves is 0.002
• The probability of selecting lesser than 6 female wolves is 0.516

The question is an illustration of binomial probability, where:

$\mathbf{P(x) = ^nC_x \times p^x \times (1 - p)^{n -x}}$

(a i) Probability of selecting 9 or more wolves out of 12, before 1918

The given parameters are:

$\mathbf{p = 0.60}$ --- the probability of selecting a male wolf

So, we have:

$\mathbf{P(x \ge 9) = P(9) + P(10) + P(11) + P(12)}$

Using $\mathbf{P(x) = ^nC_x \times p^x \times (1 - p)^{n -x}}$, we have:

$\mathbf{P(x \ge 9) = ^{12}C_9 \times 0.6^9 \times (1 - 0.6)^{12-9} +..............+^{12}C_{12} \times 0.6^{12} \times (1 - 0.6)^{12-12} }$

$\mathbf{P(x \ge 9) = 220 \times 0.00064497254 +..........+1 \times 0.00217678233}$

$\mathbf{P(x \ge 9) =0.225 }$

(a ii) Probability of selecting 9 or more female wolves

The given parameters are:

$\mathbf{p = 0.40}$ --- the probability of selecting a female wolf

So, we have:

$\mathbf{P(x \ge 9) = P(9) + P(10) + P(11) + P(12)}$

Using $\mathbf{P(x) = ^nC_x \times p^x \times (1 - p)^{n -x}}$, we have:

$\mathbf{P(x \ge 9) = ^{12}C_9 \times 0.4^9 \times (1 - 0.4)^{12-9} +..............+^{12}C_{12} \times 0.4^{12} \times (1 - 0.4)^{12-12} }$

$\mathbf{P(x \ge 9) = 220 \times 0.0000566231+..........+1 \times 0.00001677721}$

$\mathbf{P(x \ge 9) =0.015 }$

(a ii) Probability of selecting fewer than 6 female wolves

The given parameters are:

$\mathbf{p = 0.40}$ --- the probability of selecting a female wolf

Using the complement rule, we have:

$\mathbf{P(x < 6) = 1 - P(x \ge 6)}$

So, we have:

$\mathbf{P(x < 6) = 1 - [P(6) + P(7) + P(8) + P(x \ge 9)]}$

Using $\mathbf{P(x) = ^nC_x \times p^x \times (1 - p)^{n -x}}$, we have:

$\mathbf{P(x < 6) = 1 - [^{12}C_6 \times 0.4^6 \times 0.6^6 + ^{12}C_7 \times 0.4^7 \times 0.6^5 + ^{12}C_8 \times 0.4^8 \times 0.6^4 + P(x \ge 9)}$$\mathbf{P(x < 6) = 1 - [924 \times 0.00019110297 +........ + 0.0153]}$

$\mathbf{P(x < 6) = 1 - [0.335]}$

$\mathbf{P(x < 6) = 0.665}$

(b i) Probability of selecting 9 or more wolves out of 12, since 1918

The given parameters are:

$\mathbf{p = 0.70}$ --- the probability of selecting a male wolf

So, we have:

$\mathbf{P(x \ge 9) = P(9) + P(10) + P(11) + P(12)}$

Using $\mathbf{P(x) = ^nC_x \times p^x \times (1 - p)^{n -x}}$, we have:

$\mathbf{P(x \ge 9) = ^{12}C_9 \times 0.7^9 \times (1 - 0.7)^{12-9} +..............+^{12}C_{12} \times 0.7^{12} \times (1 - 0.7)^{12-12} }$

$\mathbf{P(x \ge 9) = 220 \times 0.00108954738+..........+1 \times 0.0138412872}$

$\mathbf{P(x \ge 9) =0.493 }$

(b ii) Probability of selecting 9 or more female wolves

The given parameters are:

$\mathbf{p = 0.30}$ --- the probability of selecting a female wolf

So, we have:

$\mathbf{P(x \ge 9) = P(9) + P(10) + P(11) + P(12)}$

Using $\mathbf{P(x) = ^nC_x \times p^x \times (1 - p)^{n -x}}$, we have:

$\mathbf{P(x \ge 9) = ^{12}C_9 \times 0.3^9 \times (1 - 0.3)^{12-9} +..............+^{12}C_{12} \times 0.3^{12} \times (1 - 0.3)^{12-12} }$

$\mathbf{P(x \ge 9) = 220 \times 0.00000675126+..........+1 \times 5.31441e-7}$

$\mathbf{P(x \ge 9) =0.002 }$

(b iii) Probability of selecting fewer than 6 female wolves

The given parameters are:

$\mathbf{p = 0.40}$ --- the probability of selecting a female wolf

Using the complement rule, we have:

$\mathbf{P(x < 6) = 1 - P(x \ge 6)}$

So, we have:

$\mathbf{P(x < 6) = 1 - [P(6) + P(7) + P(8) + P(x \ge 9)]}$

Using $\mathbf{P(x) = ^nC_x \times p^x \times (1 - p)^{n -x}}$, we have:

$\mathbf{P(x < 6) = 1 - [^{12}C_6 \times 0.3^6 \times 0.7^6 + ^{12}C_7 \times 0.3^7 \times 0.7^5 + ^{12}C_8 \times 0.3^8 \times 0.7^4 + P(x \ge 9)}$

$\mathbf{P(x < 6) = 1 - [924 \times 0.4^6 \times 0.7^6 + 792 \times 0.3^7 \times 0.7^5 + 495 \times 0.3^8 \times 0.7^4 + 0.002}$

$\mathbf{P(x < 6) = 1 - [0.484]}$

$\mathbf{P(x < 6) = 0.516}$

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