Executive Solution Overview
SEE ATTACHMENT – ## WEEK3.docx **## WEEK3.docx** **Conditional Probability:** What affect does a condition have on the probability? Let's compare and find out. Example: Let's look at a standard deck of 52 playing cards. No replacement: What is the probability of getting 2 kings assuming that the fir…
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SEE ATTACHMENT
– ## WEEK3.docx
**## WEEK3.docx**
**Conditional Probability:**
What affect does a condition have on the probability? Let's compare and find out.
Example: Let's look at a standard deck of 52 playing cards.
No replacement: What is the probability of getting 2 kings assuming that the first one is not replaced? (Note: There are 4 kings in a deck of cards. And a deck of cards has 52 cards.)
Answer: 4/52 * 3/51 = 1/13*1/17= 1/221
Replacement: What is the probability of getting 2 kings assuming that the first one is replaced?
Answer: 4/52 * 4/52 = 1/13 * 1/13 = 1/169
**Make up a similar example to the above problem using (a) with replacement and (b) without replacement. You may use playing cards, or a similar scenario, however be sure that you can represent both with and without replacement. Flipping a coin 10 times is not appropriate because it only represents the "without replacement" condition. Do not solve your own problem – that will be done by your classmates.**
**## WEEK3.docx**
**Conditional Probability:**
What affect does a condition have on the probability? Let's compare and find out.
Example: Let's look at a standard deck of 52 playing cards.
No replacement: What is the probability of getting 2 kings assuming that the first one is not replaced? (Note: There are 4 kings in a deck of cards. And a deck of cards has 52 cards.)
Answer: 4/52 * 3/51 = 1/13*1/17= 1/221
Replacement: What is the probability of getting 2 kings assuming that the first one is replaced?
Answer: 4/52 * 4/52 = 1/13 * 1/13 = 1/169
**Make up a similar example to the above problem using (a) with replacement and (b) without replacement. You may use playing cards, or a similar scenario, however be sure that you can represent both with and without replacement. Flipping a coin 10 times is not appropriate because it only represents the "without replacement" condition. Do not solve your own problem – that will be done by your classmates.**
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