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Use the Counting Rulesto compute Probabilities
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2 of these and 4 of those
• A classic type of problem• You have various subgroups.• When you pick 6, what is the probability that
you get 2 of this group and 4 of that group?• Jellybeans: 30 red, 30 yellow, 40 other• Choose 6. Find P(2 red and 4 yellow)
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2 red out of 30; 4 yellow out of 40
• Analysis – you must THINK! – “This is a Fundamental Counting Principle situation…– One event is drawing 2 red out of 30– The other is drawing 4 yellow out of 40– FUNDAMENTAL COUNTING PRINCIPLE says to
multiply how many ways for each of them.– Each of these events is modeled by a COMBINATION,
because the order doesn’t matter.• So how do you write it in Combination language?
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Computing the Probability
• Jellybeans: 30 red, 30 yellow, 40 other• Choose 6. Find P(2 red and 4 yellow)
• Always go back to • Numerator: • Denominator:
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“Exactly aces”
• Draw 5 cards, what is the probability of exactly 0 aces?
• We can do this with our earlier techniques:– P(first card not at ace) = ____ / 52, times …– P(second card not an ace) = ____ / 51, times …– P(third card not an ace) = ____ / 50, times …– P(fourth card not an ace) = ____ / 49, times …– P(fifth card not an ace) = ____ / 48
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“Exactly aces”
• P(0 aces out of 5 cards drawn)• A more sophisticated view– 5 non-aces out of 52 cards– How many non-aces are there?
• Numerator: ways to get 5 non-aces: • Denominator: total 5-card hands: • P(0 aces) =
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“Exactly aces”
• P(exactly 1 ace out of 5 cards drawn)• Our earlier techniques could do P(≥1 ace)• But P(=1 ace) would be harder or impossible• Counting techniques makes it easier– Choose 1 ace out of 4 aces– Choose 4 other cards out of 48 non-aces
• P(1 ace) =
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“Exactly aces”
• Similiarly for 2 aces, 3 aces, 4 aces:• P(2 aces) = • P(3 aces) = • P(4 aces) = • Check: P(0) + P(1) + P(2) + P(3) + P(4) must
total to exactly 1.000000000000000000. Why?
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Probability of a Full House
• Three of a kind– Choose 1 out of 13 ranks– Choose 3 out of 4 suits
• One pair– Choose 1 out of the remaining 12 ranks– Choose 2 out of the 4 suits
• P(full house) =
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Probability of a Flush
• A Flush: five cards all of the same suit– Choose 1 out of the 4 suits– Take 5 out of the 13 ranks
• P(flush) =
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Powerball Jackpot
• You choose 5 out of the 59 white numbers– All 5 match the 5 winners
• You choose 1 out of the 39 red numbers– And it matches the winner
• Numerator is • Denominator is possible ways to play the
ticket, not counting the extra PowerPlay “multiplier” option.
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Powerball $200,000
• You choose 5 out of the 59 white numbers– All 5 match the 5 winners
• You choose 1 out of the 39 red numbers– And it does not match the winner
• Numerator is – Notice we still have 5 out of 5 on the white
numbers– But the Powerball choice is 1 out of 38 losers
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Powerball $10,000
• You choose 5 out of the 59 white numbers– 5 winners but you picked got 4 of them– 54 losers and you picked one of those
• You choose 1 out of the 39 red numbers– And it matches the winner
• Numerator is – For the $100 via 4 white only with no red match,
just change the to a
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Powerball $7
• Two ways to win $7
• 3 white matches, 2 losers; red is no match
• Another way: 2 white matches, 3 losers, and the red powerball matches
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Two ways to lose
• Match absolutely nothing at all– 5 out of the 54 losing white numbers– 1 out of the 38 losing red powerball numbers
• Or match 1 white number only– 1 out of the 5 winning white numbers– 4 out of the 54 losing white numbers– 1 out of the 38 losing red powerball numbers