Math  /  Data & Statistics

Question16. A pen manufacturer gets its pen cartridges from 2 suppliers. 58%58 \% of the cartridges come from supplier A and 2.25%2.25 \% of them are defective. 42%42 \% of the cartridges come from supplier B and 1.75%1.75 \% of them are defective. Answer the following questions: (a) Draw a tree diagram representing the problem. (b) Find the probability that a cartridge is defective and from supplier A. (c) Find the probability that a randomly chosen cartridge is not defective.

Studdy Solution

STEP 1

What is this asking? We're investigating the probability of defective pen cartridges from two suppliers, figuring out the chances of getting a defective one from each and the overall chance of getting a non-defective cartridge. Watch out! Don't mix up percentages and probabilities – remember to convert percentages to decimals before calculating probabilities!

STEP 2

1. Draw the tree diagram
2. Probability of defective cartridge from supplier A
3. Probability of a non-defective cartridge

STEP 3

Alright, let's **visualize** this problem with a **tree diagram**!
We'll start with the **two suppliers**: Supplier A and Supplier B.

STEP 4

From Supplier A, we have two branches: **defective** and **non-defective**.
Supplier A provides 58%58\% of the cartridges, so let's write 0.580.58 on that branch.
Of those, 2.25%2.25\% are defective, so we write 0.02250.0225 on the defective branch and 10.0225=0.97751 - 0.0225 = 0.9775 on the non-defective branch.

STEP 5

Now for Supplier B!
They supply 42%42\% of the cartridges (0.420.42), with 1.75%1.75\% being defective (0.01750.0175).
So the non-defective branch from Supplier B gets 10.0175=0.98251 - 0.0175 = 0.9825.
Our tree diagram is complete!

STEP 6

We want the probability of a cartridge being both defective AND from Supplier A.
Looking at our tree diagram, we simply **multiply** the probabilities along the branches leading to this outcome.

STEP 7

P(Defective and from A)=P(A)P(DefectiveA)=0.580.0225=0.01305 P(\text{Defective and from A}) = P(\text{A}) \cdot P(\text{Defective} | \text{A}) = 0.58 \cdot 0.0225 = \mathbf{0.01305} So, there's a 0.01305\mathbf{0.01305} probability, or 1.305%\mathbf{1.305}\%, that a randomly chosen cartridge is defective and from Supplier A.

STEP 8

To find the probability of a non-defective cartridge, we need to consider both suppliers.
We'll **calculate** the probability of a non-defective cartridge from each supplier and then **add** those probabilities together.

STEP 9

From Supplier A: P(Non-defective and from A)=0.580.9775=0.56755 P(\text{Non-defective and from A}) = 0.58 \cdot 0.9775 = 0.56755

STEP 10

From Supplier B: P(Non-defective and from B)=0.420.9825=0.41265 P(\text{Non-defective and from B}) = 0.42 \cdot 0.9825 = 0.41265

STEP 11

Now, let's **add** these probabilities: P(Non-defective)=0.56755+0.41265=0.9802 P(\text{Non-defective}) = 0.56755 + 0.41265 = \mathbf{0.9802} The probability of picking a non-defective cartridge is 0.9802\mathbf{0.9802}, or 98.02%\mathbf{98.02}\%.

STEP 12

(a) The tree diagram was constructed in the steps above. (b) The probability that a cartridge is defective and from supplier A is 0.013050.01305 or 1.305%1.305\%. (c) The probability that a randomly chosen cartridge is not defective is 0.98020.9802 or 98.02%98.02\%.

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