Question

ap free response practice (past ap exam questions)
approximately 3.5 percent of all children born in a certain region are from multiple births (that is, twins, triplets, etc.). of the children born in the region who are from multiple births, 22 percent are left - handed. of the children born in the region who are from single births, 11 percent are left - handed.
(a) what is the probability that a randomly selected child born in the region is left - handed?
(b) what is the probability that a randomly selected child born in the region is a child from a multiple birth, given that the child selected is left - handed?
(c) a random sample of 20 children born in the region will be selected. what is the probability that the sample will have at least 3 children who are left - handed?

Explanation:

Part (a)

Step1: Define events and probabilities

Let \( M \) be the event of multiple births, \( S \) be the event of single births. We know \( P(M)=0.035 \), \( P(S) = 1 - 0.035=0.965 \). Let \( L \) be the event of left - handed. We know \( P(L|M) = 0.22 \), \( P(L|S)=0.11 \). We use the law of total probability: \( P(L)=P(L|M)P(M)+P(L|S)P(S) \)

Step2: Substitute values into the formula

\( P(L)=(0.22\times0.035)+(0.11\times0.965) \)
First, calculate \( 0.22\times0.035 = 0.0077 \) and \( 0.11\times0.965=0.10615 \)
Then, \( P(L)=0.0077 + 0.10615=0.11385\approx0.114 \)

Step1: Recall the formula for conditional probability

The formula for conditional probability is \( P(M|L)=\frac{P(L|M)P(M)}{P(L)} \)

Step2: Substitute the known values

We know \( P(L|M) = 0.22 \), \( P(M)=0.035 \) and from part (a) \( P(L)=0.11385 \)
\( P(M|L)=\frac{0.22\times0.035}{0.11385}=\frac{0.0077}{0.11385}\approx0.0676 \)

Step1: Identify the distribution

Let \( X \) be the number of left - handed children in the sample of \( n = 20 \) children. We first need to find the probability that a child is left - handed, \( p\). From part (a), \( p = 0.11385 \). \( X\sim B(n = 20,p=0.11385) \) (binomial distribution), where \( P(X = k)=C_{n}^{k}p^{k}(1 - p)^{n - k} \), and \( C_{n}^{k}=\frac{n!}{k!(n - k)!} \)
We want \( P(X\geq3)=1 - P(X = 0)-P(X = 1)-P(X = 2) \)

Step2: Calculate \( P(X = 0) \)

\( P(X = 0)=C_{20}^{0}(0.11385)^{0}(1 - 0.11385)^{20}=(0.88615)^{20}\approx0.108 \)

Step3: Calculate \( P(X = 1) \)

\( P(X = 1)=C_{20}^{1}(0.11385)^{1}(0.88615)^{19}=20\times0.11385\times(0.88615)^{19} \)
\( (0.88615)^{19}\approx0.1218 \), so \( P(X = 1)=20\times0.11385\times0.1218\approx0.275 \)

Step4: Calculate \( P(X = 2) \)

\( P(X = 2)=C_{20}^{2}(0.11385)^{2}(0.88615)^{18}=\frac{20!}{2!18!}\times(0.11385)^{2}\times(0.88615)^{18} \)
\( \frac{20!}{2!18!}=\frac{20\times19}{2}=190 \), \( (0.11385)^{2}\approx0.01296 \), \( (0.88615)^{18}\approx0.1374 \)
\( P(X = 2)=190\times0.01296\times0.1374\approx0.334 \)

Step5: Calculate \( P(X\geq3) \)

\( P(X\geq3)=1-(0.108 + 0.275+0.334)=1 - 0.717 = 0.283 \)
(Note: The values of the powers can also be calculated using a calculator or software for more accuracy. Another way is to use the binomial probability formula in a calculator - like approach. The approximate value may vary slightly depending on the method of calculation, but the general approach is as above.)

Answer:

The probability is approximately \( 0.114 \)

Part (b)