# Math Help - probability using die and deck of cards? help!

1. ## probability using die and deck of cards? help!

At this current moment, I am completely stuck. I am currently in Business Statistics and have been thrown a last minute project that I just dont get?

I need help with number 1, 2 and 3(d).

Im not looking for the answers but for someone to please explain to me how to get started...Thank you so very much.
• 1. You are to use one (1) die and cast it 8 times. Record each outcome. After you have recorded your outcomes, compute the probability of the series of outcomes you encountered being duplicated.

• 2. Using a 2-sided coin, toss the coin 4 times. Record your outcomes then compute the probability of experiencing 4 heads against the probability of your recorded outcomes.

• 3. Use your deck of playing cards. Remove the Jokers. (Record each draw)

• A. What is the probability that you will draw the Ace of Hearts on your first try?

• B. Whatever card you drew in Part A, return it to the deck. What is the probability that you will draw that same card on your second draw?

• C. Do not replace the second card drawn. What is the probability you will draw a card from the same suit on your third draw?

• D. What is the probability (without replacing any draw) that you will draw 5 cards from the same suit in 5 consecutive draws?

2. Hello, oliviashea!

Here's 3(D) . . .

3. Use your deck of playing cards. Remove the Jokers.

D. What is the probability (without replaciement) that you will draw 5 cards from the same suit
in 5 consecutive draws?
The 1st card can be any of the 52 cards: . $\frac{52}{52} = 1$

The 2nd card must be one of the 12 remaining cards of the same suit: . $\frac{12}{51}$

The 3rd card must be one of the 11 remaining cards of the same suit: . $\frac{11}{50}$

The 4th card must be one of the 10 remaining cards of the same suit: . $\frac{10}{49}$

The 5th card must be one of the 9 remaining cards of the same suit: . $\frac{9}{48}$

Therefore: . $P(\text{flush}) \:=\:1\cdot\frac{12}{51}\cdot\frac{11}{50}\cdot\fr ac{10}{49}\cdot\frac{9}{48} \:=\:\frac{33}{16,\!660}$