Question

Four cards are to be drawn simultaneously from a deck of 52. Find the probability that

*(No clue if any of these are right)

(a). all four are clubs

13/52 * 12/51 * 11/ 50 * 10/ 51 = 0.00264

(b) at least one is not a club

(39 * 51 * 50 * 49 ) / (52 * 51 * 50 * 49) = 0.94

(c) three are black and one is red

(21 * 20 * 19 * 21) / (52 * 51 * 50 * 49) = 0.0257

(d) exactly two are picture card

- I have looked up a few examples of this specific problem and it shows using combinations. I just do not understand how to tell when you are suppose to use a combination.

(e) either two or three of the four cards are picture cards

- Again, I have looked up a few examples of this specific problem and it show using combinations, I just do not understand how to tell when to use them.

Answer 1

PAGE No. гетс / ! soicerion, giren that four cords to be down simulroneously from dece of 52. . Find the prob. all fours coe Club) * There, 13 Clubs cords in deck of 52 → P(al fours aveclub) = 136 4 52c4 13x1211)O 52X51X50X49 0.00264) 2. A Poob that atleast one is not a club. → PCatleast one is not club) = 1- PCall fouos ore Club) = 1 -0.00264/ o 987359 0997359 3. poob(Thore blocondone Jed) In deck of 52, 26 - block cords 26 - Red cords These foori PC Three blocks and One ded) = 26CSACI 5704

PAGE NO. DATE 26*25*24 26 3x 201 52XSIX SOX49 6x3x2XT 0.249699. 4. Poob C EKOCHy too are pictures coods) There are 3 pictures coods in each of swers (Diamond, club beool, Spode) Therefore 3*4 = 12 pictures coods > P Exochy troo ooe picrede coods) 1462 X 52€ 12 C 2 x 40c2 52C2 - = 0.190156 of the four coods 5. Either two of three one picture coods. P( Either reco 07 thee of four picture cords) - 1262*4062 + 1263 x 3941 0.22 1848