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

Bunuel chetan2u Please share (simplified!) OE

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

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