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crimeas [40]
3 years ago
6

A bag has 8 number tiles numbered one through eight. You draw a number tile, replace the tile and draw a second tile. What is th

e probability that you will draw a 6 and then a 7? Only answer if you answer is 100% correct.
A. 1/64
B. 1/56
C. 1/16
D. 1/8
Mathematics
2 answers:
k0ka [10]3 years ago
8 0
Total number tiles = 8
Number of 6 = 1
Number of 7 = 1

P(a 6 and then a 7) = (1/8)(1/8) = 1/64 (Answer A)
solmaris [256]3 years ago
5 0
The answer is D because they take a tile out which makes it 7 tiles in the bag but then they put it back which makes the bag have 8 tiles in it again.
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b) How many ways can you deal cards (from a deck of 52) to 4 people when each player gets 7 cards. 2 hidden and 5 visible. Assum
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{52 \choose 7}{45 \choose 7}{38 \choose 7}{31 \choose 7}{7 \choose 2}^4

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The dealing of the cards can be seen in the following way:

We first have to choose which 7 cards are going to be dealt to the 1st player. So we have to pick 7 cards out of the 52 available cards. This can be done in {52 \choose 7} ways. Now that we have chosen which 7 cards are going to be dealt to the 1st player, we have to choose which 2 of them are going to be the hidden ones. So we have to pick 2 cards out of the 7 cards to be the hidden ones. This can be done in {7 \choose 2} ways. At this point we now know which cards are being dealt to the 1st player, and which ones are hidden for him. Then we have to choose which 7 cards to deal to the 2nd player, out of the remaining ones. So we have to pick 7 out of 52-7=45. (since 7 have been already dealt to the 1st player). This can be done in {45 \choose 7}. Then again, we have to pick which 2 are going to be the hidden cards for this 2nd player. So we have to pick 2 out of the 7. This can be done in {7 \choose 2} ways. Then we continue with the 3rd player. We have to choose 7 cards out of the remaining ones, which at this point are 52-7-7=38. This can be done in {38 \choose 7}. And again, we have to choose which ones are the hidden ones, which can be chosen in {7 \choose 2} ways. Finally, for the last player, we choose 7 out of the remaining cards, which are 52-7-7-7=31. This can be done in {31 \choose 7} ways. And choosing which ones are the hidden ones for this player can be done in {7 \choose 2} ways. At the end, we should multiply all our available choices on each step, to get the total choices or total ways to deal the cards to our 4 players (since dealing the cards is a process of several steps with many choices on each step).

{52 \choose 7}{7 \choose 2}{45 \choose 7}{7 \choose 2}{38 \choose 7}{7 \choose 2}{31 \choose 7}{7 \choose 2}={52 \choose 7}{45 \choose 7}{38 \choose 7}{31 \choose 7}{7 \choose 2}^4

7 0
4 years ago
3.8 repeating as a fraction
tester [92]
Ans=3.77777777...
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