Question

1. approximately 3.5 percent of all children born in a certain region are from multiple births (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. draw a tree diagram to model this scenario.
b. what is the probability that a randomly selected child born in the region is left - handed? show your work.
c. 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? show your work.

Explanation:

Part A: Tree Diagram Explanation

To draw the tree diagram, we start with the first level representing the type of birth (multiple or single). The probability of multiple births (\(M\)) is \(0.035\), so the probability of single births (\(S\)) is \(1 - 0.035 = 0.965\). From each of these branches, we add a second level for being left - handed (\(L\)) or right - handed (not \(L\)). For multiple births, \(P(L|M)=0.22\) and \(P(\text{not }L|M)=1 - 0.22 = 0.78\). For single births, \(P(L|S)=0.11\) and \(P(\text{not }L|S)=1 - 0.11 = 0.89\).

Part B: Probability of being left - handed

We use the law of total probability. The law of total probability states that if we have a partition of the sample space (here, the partition is multiple births \(M\) and single births \(S\)), then \(P(L)=P(L|M)P(M)+P(L|S)P(S)\)

Step 1: Identify the probabilities

We know that \(P(M) = 0.035\), \(P(L|M)=0.22\), \(P(S)=0.965\), and \(P(L|S)=0.11\)

Step 2: Apply the law of total probability

$$
LATEXBLOCK0
$$

Part C: Probability of multiple births given left - handed

We use Bayes' theorem. Bayes' theorem states that \(P(M|L)=\frac{P(L|M)P(M)}{P(L)}\)

Step 1: Identify the values

We already found that \(P(L|M) = 0.22\), \(P(M)=0.035\), and \(P(L)=0.11385\) from part B.

Step 2: Apply Bayes' theorem

$$
LATEXBLOCK1
$$

Part B Answer

The probability that a randomly selected child is left - handed is \(0.11385\)

Part C Answer

The probability that a randomly selected left - handed child is from a multiple birth is approximately \(0.0676\)

Answer:

Part A: Tree Diagram Explanation

To draw the tree diagram, we start with the first level representing the type of birth (multiple or single). The probability of multiple births (\(M\)) is \(0.035\), so the probability of single births (\(S\)) is \(1 - 0.035 = 0.965\). From each of these branches, we add a second level for being left - handed (\(L\)) or right - handed (not \(L\)). For multiple births, \(P(L|M)=0.22\) and \(P(\text{not }L|M)=1 - 0.22 = 0.78\). For single births, \(P(L|S)=0.11\) and \(P(\text{not }L|S)=1 - 0.11 = 0.89\).

Part B: Probability of being left - handed

We use the law of total probability. The law of total probability states that if we have a partition of the sample space (here, the partition is multiple births \(M\) and single births \(S\)), then \(P(L)=P(L|M)P(M)+P(L|S)P(S)\)

Step 1: Identify the probabilities

We know that \(P(M) = 0.035\), \(P(L|M)=0.22\), \(P(S)=0.965\), and \(P(L|S)=0.11\)

Step 2: Apply the law of total probability

$$
LATEXBLOCK0
$$

Part C: Probability of multiple births given left - handed

We use Bayes' theorem. Bayes' theorem states that \(P(M|L)=\frac{P(L|M)P(M)}{P(L)}\)

Step 1: Identify the values

We already found that \(P(L|M) = 0.22\), \(P(M)=0.035\), and \(P(L)=0.11385\) from part B.

Step 2: Apply Bayes' theorem

$$
LATEXBLOCK1
$$

Part B Answer

The probability that a randomly selected child is left - handed is \(0.11385\)

Part C Answer

The probability that a randomly selected left - handed child is from a multiple birth is approximately \(0.0676\)