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# While staying in a 15-story hotel, Polya plays the following game. She enters an elevator on the $6^{\mathrm{th}}$ floor. She flips a fair c

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While staying in a 15-story hotel, Polya plays the following game. She enters an elevator on the 6th floor. She flips a fair coin five times to determine her next five stops. Each time she flips heads, she goes up one floor. Each time she flips tails, she goes down one floor. What is the probability that each of her next five stops is on the 7th floor or higher? Express your answer as a common fraction.

3rd attempt, i need a valid solution with is not posted before, i still have no idea how to solve this

Feb 2, 2018

#1
+111389
0

...........

Feb 2, 2018
edited by CPhill  Feb 2, 2018
#2
+111389
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Here's my reasoning......

1st Flip....possible scenarios

( Starting Floor   ⇒ Ending Floor )

6 ⇒ 7

6 ⇒ 5

Second Flip ....possible scenarios

7 ⇒ 8

7 ⇒ 6

5 ⇒ 6

5 ⇒ 4

Third flip - possible scenarios

8 ⇒ 9

8 ⇒ 7

6 ⇒ 7     [ this one is no good...being at the 6th floor intially means that she was below the 7th floor on one of the flips....the requirement is being at 7 or higher on all 5 flips ]

6 ⇒ 5

4 ⇒ 5

4 ⇒ 3

Fourth Flip - possible scenarios

9 ⇒ 10         5 ⇒ 6

9 ⇒ 8           5 ⇒ 4

7 ⇒ 8           3 ⇒ 4

7 ⇒ 6           3 ⇒ 2

Last Flip - possible scenarios

10 ⇒ 11       6 ⇒ 7       2 ⇒ 3

10 ⇒ 9         6 ⇒ 5       2 ⇒ 1

8 ⇒ 9           4 ⇒ 5

8 ⇒ 7           4 ⇒ 3

Total possible scenarios  =   2 + 4 + 6 + 8 + 10   =  30

Notice, however, that only 11 of these guarantee a floor of 7th or higher on each of  5 flips

Favorable flip 1 outcomes

6 ⇒ 7

Favorable  flip 2 outcomes

7 ⇒ 8

Favorable Flip 3 outcomes

8 ⇒ 9

8 ⇒ 7

Favorable Flip 4 outcomes

9 ⇒10

9 ⇒ 8

7 ⇒ 8

Favorable Flip 5 outcomes

10 ⇒ 11

10 ⇒ 9

8 ⇒ 9

8 ⇒ 7

So.......the probability  that she is on a floor 7th or higher on each of the 5 flips  is

11 / 30

Feb 2, 2018
#3
+473
+1

Sorry but that is incorrect

RektTheNoob  Feb 5, 2018