In a bag, there are three counters marked with the digit "3" and five

Question

In a bag, there are three counters marked with the digit "3" and five counters marked with digit "2". Four counters are drawn out of the bag one at a time with replacement after each draw, find the probability that the sum of the digits on the counters is more than or equal to 10?

A. 621/4096

B. 441/4096

C. 423/2048

D. 1971/4096

E. 2125/4096

A. 621/4096

B. 441/4096

C. 423/2048

D. 1971/4096

E. 2125/4096

IanStewart · Answer

If we get a sum less than 10, we either pick four 2's, or we pick one 3 and three 2's.

Since we're picking with replacement, the probability we pick four 2's is (5/8)^4

The probability we pick 3222 in that order is (3/8)(5/8)^3, but we can pick one 3 and three 2s in four different orders (the 3 can be picked first, second, third, or fourth), so the probability we get exactly one 3 is (4)(3/8)(5/8)^3

So the probability we do *not* get the result the question asks for is

(5/8)^4 + (4)(3/8)(5/8)^3 = 5^3(5 + 4*3) /8^4 = 17*5^3/8^4

and subtracting from 1, we'll find the probability we do get the result the question asks for, so the answer is

1 -   =  /8^4

Now the numerator is odd, but the only prime divisor of the denominator is 2, so this fraction can't cancel, and 8^4 = 4096 must be our denominator. The units digit of 17*5^3 must be 5, so the units digit of 8^4 - (17)(5^3) must be 1, and since (17)(5^3) is fairly close to 2000, the only reasonable answer is D.

believerinthy · Answer

I'm getting it from another way and I get a figure of 1764/4096.

Is it fairly reasonable to pick D?

wishmasterdj · Answer

Bunuel   chetan2u  Please share (simplified!) OE

chetan2u · Answer

wishmasterdj

We have three of 3s and five of 2s. When we pick four with replacement, the total should be equal to or greater than 10.
'With replacement' means the set of these  'three of 3s and five of 2s'  will be available for each pick.

The different ways in which total is equal to or more than 10 are

(1) All four picked up are 3s. 
ways to pick 3 in each pick is 3 as we have three of 3s => 3*3*3*3=81
The arrangement is just 1

(2) Three picked up are 3s and one is 2. 
ways to pick 3 in each pick is 3 as we have three of 3s, and ways to pick 2 in each pick is 5 as we have five of 2s => 3*3*3*5
The arrangement is the same as arranging 4 items out of which 3 are of one kind =  \frac{4!}{3!}=4 
Total ways = 3*3*3*5*4=540

(3) Two picked up are 3s and two are 2s. 
ways to pick 3 in each pick is 3 as we have three of 3s, and ways to pick 2 in each pick is 5 as we have five of 2s => 3*3*5*5
The arrangement is the same as arranging 4 items out of which 2 are of one kind, and the other two are of similar kind=  \frac{4!}{2!2!}=6 
Total ways = 3*3*5*5*6=1350

No more possibilities as if we have three of 2s and one of 3s, we get 2+2+2+3=9<10.

So total ways that  sum\geq{10}  are  81+540+1350=1971

However, total ways to pick 4 out of 8 with replacement = 8*8*8*8=4096

P= \frac{1971}{4096}

D

Kinshook · Answer

Given: In a bag, there are three counters marked with the digit "3" and five counters marked with digit "2".
Asked: Four counters are drawn out of the bag one at a time with replacement after each draw, find the probability that the sum of the digits on the counters is more than or equal to 10?

Sum less than 10 is possible when all 4 2s, 3 2s & 1 3s are drawn.

The probability that the sum of the digits on the counters is less than 10 = (5/8)^4 + 4C1*(5/8)^3(3/8) = (5/8)^3 { 5/8 + 4*3/8 } = (5/8)^3 {5/8 + 12/8 } = (5/8)^3 {17/8} = 2125/4096
The probability that the sum of the digits on the counters is more than or equal to 10 = 1 - 2125/4096 = 1971/4096

IMO D

In a bag, there are three counters marked with the digit "3" and five

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