1. There are (52 choose 5) possible 5 card hands. How many of these hands have only one suit? 4 choices for the suit, and (13 choose 5) choices for the cards. So P(E)=4*(13 choose 5)/(52 choose 5).
For P(F), we find instead the probability of having NO hearts, which is (39 choose 5)/(52 choose 5), and then subtract from 1: P(F)=1-(39 choose 5)/(52 choose 5).
For P(E \intersect F) how many hands have at least one heart, and have all of the cards of the same suit? We must have only hearts, so (13 choose 5)/(52 choose 5).
We are given that our hand has at least one heart, we know from the P(F) calculation that there are (52 choose 5)-(39 choose 5) such hands. How many of these hands have all cards of the same suit? (13 choose 5) since the suit must be hearts. So P(E|F)=(13 choose 5)/((52 choose 5)-(39 choose 5)).
2. (10 choose 6)*(0.60)^6*(0.40)^4=about 25%.
3.No. Certainly if E occurs then F must occur, they are not independent. P(E)= 10/36, P(F)= 18/36=1/2, P(E \intersect F)=10/36 which is not equal to 10/36*1/2.
4. P(QUEEN)=1/13, P(NOTQUEEN)=12/13. P(QUEEN and FACECARD)=1/13, P(QUEEN and NOTFACECARD)=0, P(NOTQUEEN and FACECARD)=2/13, P(NOTQUEEN and NOTFACECARD)=10/13,so
P(FACECARD|NOTQUEEN)=2/13/(12/13)=1/6.
1. Let's find the probability that I win the bet. I have (9 choose 5)=126 possible choices for my set of 5 numbers. Suppose that your numbers are 1,2,3,4. I win if I guess all 4 correct: (5 choose 1)=5 ways to do this. I win if I guess exactly 3 of your numbers: (4 choose 3)*(5 choose 2)=4*10=40 ways. I win if I guess exactly 2 of your numbers (4 choose 2)*(5 choose 3)=6*10=60, so I win 105/126 =about 83%. So you had better not bet with me on this game.
2. There are 100 rolls possible. Since 10 of them are doubles, P(E)=10/100.
How many ways can we roll a 10? 9 ways: 19,28,37,46,55,64,73,82,91, so P(F)= 9/100. Since P(E\union F)=P(E)+P(F)-P(E \intersect F), we must find P(E \intersect F). How many of the doubles rolls sum to 10? Only one, 55, so P(E \intersect F)=1/100.
For P(E|F) how many of the 9 rolls of ten are doubles? Only 1 so P(E|F)=1/9.
For P(F|E) how many of the 10 double rolls are 10: only 55, so P(F|E)=1/10.
3. Let's roll one 6 sided die. Let E= 1 appears, F=2 appears. P(E|F)=0.
4. Either no defective computers (0.95)^20=0.214 or exactly one defective computer, (30 choose 1)*(0.95)^29*(0.05)=0.339. The sum is about 0.553.
5. There are 8! ways to permute one line. How many ways have 6 couples matched? (8 choose 6) (choose the couples), then switch the remaining two couples. (8 choose 6)/8!=0.000694444.
1. For exactly 3 of each, (6 choose 3)*(1/2)^3*(1/2)^3=20/64=about 31%
2. There are (8 choose 4) possible committees. How many of these committees have Cosmo and Osmo together? Just choose 2 student for the rest of the Cosmo-Osmo committee (6 choose 2). So the answer is
(6 choose 2)/(8 choose 4)=15/70.
3. There are 6*6*6=216 possible rolls which are equally likely.
For P(E) we have 6*5*4=120 possible rolls so P(E)=120/216.
For P(F) we have 5*5*5=125 possible rolls so P(F)=125/216.
For P(E \union F)=P(E)+P(F)-P(E \interesect F), must find the number of rolls which have no sixes and are different: 5*4*3=60. So P(E \union F)=120/216+125/216-60/216=175/216.
For P(E|F), we already know that F occurs 125 times. How many times do both E and F occur? We just did that- 60 times. So P(E|F)=60/125.
4. P(E)=2/(52 choose 4). If we say that no kings are drawn, we expect this number to slightly increase, since there many fewer choices available (48 cards instead of 52 to choose from) and we still have two hands possible.
1. P(brown eyed)=9/25, P(female)=12/25, P(brown eyed and female)=4/25=16% which is not equal to 9/25*12/25=17% (close though).
2. There are (7 choose 3)*4!=840 arrangements of PRECEDE.
(a) Think of the three E's as glued together, there are then 5! ways to arrange P, R, C, D, EEE, we get 5!/840=1/7.
(b) There are (6 choose 3)*3!=120 ways to arrange PRECEDE if R is the 3rd letter. How many of these ways have 3 E's in a row? The E's must be either in positions 4,5,6 or 5,6,7 (2 choices) Then permute the P C, D in 3!=6 ways, to get 2*6/120.
3. (a) Since 1+2+3+...+8=36, 7 is rolled with probability 7/36.
(b) We need either 78 or 87. What are the probs of each case? 78 is 7/36*8/36, while 87 is 8/36*7/36, so the sum is 2*7*8/36^2= about 8.6%.
4. There are (52 choose 4) possible hands. How many of them have three of a kind? Choose the number which is the three of a kind (13 ways), then choose the three cards of that number (4 choose 3), and choose another card (48 ways). We get 13*(4 choose 3)*48/(52 choose 4)= about 0.9%.
5. P(1st flip is H)=1/2. P(exactly 2 H's in four flips)=(4 choose 2)*(1/2)^4=6/16. But P(1st flip is H and exactly 2H's in four flips)=3/16, since there are only three ways to do it: HHTT, HTHT, HTTH. Since the prob of the intersection is the product of the probs, they are independent events.