Probability of games to require modification after an update

Question

After an update, out of the games on the market, 10% of them will need to be completely modified, 30% will be need to be modified partly, and 60% will require no modification. A game developer currently has 6 games on the market. What is the probability that: i) At least 3 of his games will need to be modified partly and ii) Two require a complete modification and one wont require any modifications.

For the first part, I said that the probability of at least 3 games to require partial modification is $1-$ probability of none, 1 or two to not require partial modification. I got $1-\left(0.7^6+0.7^5\cdot 0.3+0.7^4\cdot 0.3^2\right) = 0.81$ . Was I correct in my assumption?

For the second part, I just did $0.1^2\cdot 0.6\cdot 0.3^3 = 0.00016$ since I thought that the probability is just the product of each event's probability(as in, probability of no modify^2 times probability of no modify times probability of partial modification^3), but the result I get is very low and I don't know if what I did was correct or not.

Answer 1

I got $1−(0.7^6 + 0.7^5⋅0.3+0.7^4⋅0.3^2)=0.81$ . Was I correct in my assumption?

No, but you have the right idea.

The probability that exactly one of the six requires partial-modification is $\binom 61\cdot 0.7^5\cdotp 0.3$, because it can be any one of the six, not just the last one.

Likewise the probability that exactly two of the six requires partial-modification is $\binom 62\cdot 0.7^4\cdotp 0.3^2$, because it can be any two of the six, not just the last two.

$$1−(0.7^6 + 6\cdotp 0.7^5\cdotp 0.3+15\cdotp 0.7^4\cdotp 0.3^2)=0.25569$$

Alternatively: $20\cdotp 0.7^3\cdotp0.3^3+15\cdotp 0.7^2\cdotp 0.3^4+6\cdotp 0.7^1\cdotp 0.3^5+0.3^6 = 0.25569$

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Apply the same principle to the second problem.