How to Play Poker Videos

Who can solve these problems about probability?

1.A pair of dice is tossed. Find the probability of getting a) a total of 8? B) at most a total of 5? 2.Two cards are drawn in succession from a deck without replacement. What is the probability that both cards are greater than 2 and less than 8? 3.If each coded item in catalog begins with 5 different letters and 3 distinct nonzero digits, find the probability of randomly selection one of these coded items with the first letter a vowel and last digit even? 4.In a high school graduating class of 100 students, 54 studied mathematics, 69 studied history, and 35 studied both mathematics and history. If one of these students is selected at random, find the probability that the student took history but not mathematics? 5.In a poker hand consisting of cards, find he probability of holding 3 aces? 6.In a game of poker where a player is dealt 5 cards from a deck of 53 playing cards (52 cards + 1 joker), what is the probability of getting a full house?

Public Comments

1. 1. Total of 8: 5/36 5 or less: 10/36 2. 20/52 first draw 19/51 second draw Multiply together 380/2652 = 14.33% 3. Don't know what exactly is going on here. 4. These numbers don't add up. If 35 studied both math and history, then (54-35) 19 studied mathematics only. Then you have (69-35) 34 studying only History. 34 for history + 19 for math + 35 for both = 88, not 100. Check your numbers and if you didn't make a mistake copying then I believe your professor has made a mistake.
2. E= R/N E is your target event. R is the total number of ways that event can be achieved N is the total number of UNIQUE events that can be achieved 1a) E=8 or 2+6,3+5,4+4 R=3 N=21 NOT 36 21 because: 1+1, 1+2, 1+3, 1+4, 1+5, 1+6 = 6 UNIQUE events 2+2, 2+3, 2+4, 2+5, 2+6 = 5 UNIQUE events 3+3, 3+4, 3+5, 3+6 = 4 UNIQUE events 4+4, 4+5, 4+6 = 3 UNIQUE events 5+5, 5+6 = 2 UNIQUE events 6+6 = 1 UNIQUE event 6 + 5 + 4 + 3 + 2 + 1 = 21 TOTAL Unique events possible Probability is 1/7 or 0.142857 or 14.2857% for rolling an eight. 1b) By saying "at most a total of 5" the assumption I am making is that totals of 2, 3, 4 & 5 are the events. E =< 5 R = 6 N = 21 Answer is 6/21 or 0.2857 or 28.57% of being a total of 5 or less. 2) Assuming no jokers and assuming that 2's and below and 8's and above are NOT included. First card: 2 < E < 8 R = 20 = 52 - (4 A's + 4 2's + 4 8's + 4 9's + 4 10's + 4 J's + 4 Q's + 4 K's) N = 52 Probability: 20/52 or 0.3846 or 38.46% Second card would have 51 cards to pick from, ergo: Assuming card 1 met criteria: 19/51 or 0.3725 or 37.25% If card 1 met criteria: 0.3846 * 0.3725 = 0.14326 or 14.326% chance of getting two consecutive cards that met the conditions. 3) Starting with AAAAA000 and going through to ZZZZZ999 there are a possible 11,881,376,000 combinations. 26 letters in first position * 26 letters in second position * 26 letters in third position * 26 letters in fourth position * 26 letters in fifth position * 1000 different numbers for each letter set For R then, there are a possible 1,142,440,000 possible codes that meet the condition of a vowel in the first letter position and an even number in the last letter position. 5 vowels in first position * 26 letters in second position * 26 letters in third position * 26 letters in fourth position * 26 letters in fifth position * 500 numbers that are even for each letter set Probability: 1,142,440,000 / 11,881,376,000 or 0.09615 or 9.615% 4) 69 students took history of which 35 took math as well. So, 69 - 35 = 34 students that took history alone. E = history alone R = 34 N = 100 Probability: 34/100 or 0.34 or 34% chance of selecting a student that took history alone. 5 ) bragging rights just aren't worth it any more.... LOL, too many questions! I surrender, gonna go gaming. Hope those answers helped out a bit! :o)