Use the Counting Rules to compute Probabilities

1. Use the Counting Rulesto compute Probabilities

2. 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)

3. 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?

4. 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:

5. “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

6. “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) =

7. “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) =

8. “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?

9. 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) =

10. 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) =

11. 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.

12. 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

13. 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

14. 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

15. 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