**Probability with a Deck of Cards**When you draw two cards from a standard deck of 52 cards, and once the first card is drawn, it is not put back, you can determine specific probabilities as follows:a.) What is the probability of drawing a red card on the first draw and a red card on the second draw?b.) What is the probability of drawing a 7 on the first card and a King on the second card?### Explanation with Solutions:**a.) Probability of drawing a red card on the first draw and a red card on the second draw:**1. There are 26 red cards in a deck of 52 cards.2. Probability of drawing a red card first: $ P(\text{Red on first draw}) = \frac{26}{52} = \frac{1}{2} $3. After drawing the first red card, 25 red cards remain out of 51 total cards.4. Probability of drawing another red card: $ P(\text{Red on second draw | Red on first draw}) = \frac{25}{51} $5. Therefore, the combined probability: \[ P(\text{Red on first and Red on second}) = P(\text{Red on first}) \times P(\text{Red on second | Red on first}) = \frac{1}{2} \times \frac{25}{51} = \frac{25}{102} \]**b.) Probability of drawing a 7 on the first card and a King on the second card:**1. There are 4 sevens in a deck of 52 cards.2. Probability of drawing a seven first: $ P(7 \text{ on first draw}) = \frac{4}{52} = \frac{1}{13} $3. After drawing the seven, we have 4 Kings remaining out of 51 total cards.4. Probability of drawing a King: $ P(\text{King on second draw | 7 on first draw}) = \frac{4}{51} $5. Therefore, the combined probability: \[ P(7 \text{ on first and King on second}) = P(7 \text{ on first}) \times P(\text{King on second | 7 on first}) = \frac{1}{13} \times \frac{4}{51

The number of cards in a deck of cards is 52.

You draw two cards from a standard deck of 52 cards, once the first card is drawn it is not put back a.) Find P(red on the first draw and red on the second draw) = b.) Find P(7 on the first card and a King on the second card)=

You draw two cards from a standard deck of 52 cards, once the first card is drawn it is not put back a.) Find P(red on the first draw and red on the second draw) = b.) Find P(7 on the first card and a King on the second card)=

Algebra and Trigonometry (MindTap Course List)

4th Edition

ISBN:9781305071742

Author:James Stewart, Lothar Redlin, Saleem Watson

Publisher:James Stewart, Lothar Redlin, Saleem Watson

Chapter14: Counting And Probability

Section14.FOM: Focus On Modeling: The Monte Carlo Method

Problem 3P: Dividing a JackpotA game between two players consists of tossing a coin. Player A gets a point if...

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Dividing a JackpotA game between two players consists of tossing a coin. Player A gets a point if the coin shows heads, and player B gets a point if it shows tails. The first player to get six points wins an 8,000 jackpot. As it happens, the police raid the place when player A has five points and B has three points. After everyone has calmed down, how should the jackpot be divided between the two players? In other words, what is the probability of A winning and that of B winning if the game were to continue? The French Mathematician Pascal and Fermat corresponded about this problem, and both came to the same correct calculations though by very different reasonings. Their friend Roberval disagreed with both of them. He argued that player A has probability 34 of winning, because the game can end in the four ways H, TH, TTH, TTT and in three of these, A wins. Robervals reasoning was wrong. a Continue the game from the point at which it was interrupted, using either a coin or a modeling program. Perform the experiment 80 or more times, and estimate the probability that player A wins. bCalculate the probability that player A wins. Compare with your estimate from part a.
Dividing a Jackpot A game between two pIayers consists of tossing coin. Player A gets a point if the coin shows heads, and player B gets a point if it shows tails. The first player to get six points wins an $8000 jackpot. As it happens, the police raid the place when player A has five points and B has three points. After everyone has calmed down, how should the jackpot be divided between the two players? In other words, what is the probability of A winning (and that of B winning) if the game were to continue? The French mathematicians Pascal and Fermat corresponded about this problem, and both came to the same correct conclusion (though by very different reasoning's). Their friend Roberval disagreed with both of them. He argued that player A has probability of Winning, because the game can end in the four ways H, TH, TTH, TTT, and in three of these, A wins. Roberval’s reasoning was wrong. Continue the game from the point at which it was interrupted, using either a coin or a modeling program. Perform this experiment 80 or more times, and estimate the probability that player A wins. Calculate the probability that player A wins. Compare with your estimate from part (a).