A simplified roulette wheel contains only black and red pockets

Question

A simplified roulette wheel contains only black and red pockets. When spun, this roulette wheel stops, on average, at black on half of all spins and at red on half of all spins. If the wheel is spun 7 times, what is the probability that the wheel will stop at red fewer than 6 times?

A) 1/16

B) 3/8

C) 57/64

D) 15/16

E) 31/32

A) 1/16

B) 3/8

C) 57/64

D) 15/16

E) 31/32

chetan2u · Answer

Easier way would be to find probability of 6 and 7 times..
1) 7 times -- (\frac{1}{2})^7 
2) 6 times -- 7C1(\frac{1}{2})^6*(\frac{1}{2})^1 
Combined 6 or 7 times =  8*(\frac{1}{2})^7

So, fewer than 6 times = 1- 8*(\frac{1}{2})^7 =1- (\frac{1}{2})^4 = 1-\frac{1}{16} = \frac{15}{16}

D

Archit3110 · Answer

P for total < 6 times ; (1/2)^5 ; 1/32
for both red & black ; 1/32*2c1 ; 1/16
1-1/16 ; 15/16
IMO D

philipssonicare · Answer

What is the question source?

"fewer" should be treated the same as "at least" and use 1-P

ccheryn · Answer

Hi Archit ,( hope ur preparation is going great) could you explain your method . actually ur ans is spot on. but i cant understand ur explanation archit.

ccheryn · Answer

i think fewer cannot be treated as at least . do you mean, fewer = 1- atleast (p). if so then its fine.

Archit3110 · Answer

A simplified roulette wheel contains only black and red pockets. When spun, this roulette wheel stops, on average, at black on half of all spins and at red on half of all spins. If the wheel is spun 7 times, what is the probability that the wheel will stop at red fewer than 6 times?

A) 1/16

B) 3/8

C) 57/64

D) 15/16

E) 31/32

Show Answer

Official Answer

D

.. coming to your question given in question roulette wheel contains only black and red pockets ; When spun, this roulette wheel stops, on average, at black on half of all spins and at red on half of all spins. it means that we have 1/2 chance for both red & black at each spin. Now, If the wheel is spun 7 times, what is the probability that the wheel will stop at red fewer than 6 times ? fewer than 6 times would be at <=5 the total fair possibilities of getting red or black in 5 spins ; (1/2)^5 = 1/32 but whether the spin stops at black or red can be determined ; 2c1 ways so total ways ; 1/32*2c1 ; 1/16 ( black and red each) question says ; wheel will stop at red fewer than 6 times ; <=5 1-1/16 ; 15/16 ANOTHER way ; we know total fair ways are 7 so (1/2)^7 = 1/128 and for each black or red ; 1/128*2c1 ; 1/64 ( total ways) so since we need to find the probability that the wheel will stop at red fewer than 6 times i.e <=5 times we have fair chance of 2^2 ; 4 of getting red >=5 ; hence P of getting red >=5 ; 4/64 ; 1/16 and P of getting <=5 ; 1-1/16 ; 15/16 Hope this helps

queensbridge · Answer

Hi Chetan, could you please explain step 2 in your solution? Why do we use 7C1 and how does it become 8 *(1/2)^7? Thanks

ccheryn · Answer

for 6 times, ie 6R and 1B can be written as (RRRRRRB). These can be arranged in 7!/6! ways or 7c1 ways. thus 7c1 , for probability of red (1/2)^6 , for probability of black ( 1/2)^1 ( thus step 2 is formed)

3rd step is adding first two steps ( 1/2)^7 ( 1 +7) which gives = 1/2^7 * 8

hope it clarifies your doubt.

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A simplified roulette wheel contains only black and red pockets

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