Dixon wrote:

I understand how they calculated this, but here's what I don't get: say for one prime digit scenario, the solution calculates different arrangements, as in prob of 4*(1/2)^4. I get the 1/2^4 as being the probability of getting one prime number and three non-prime numbers. But why take into account the arrangements by multiplying by 4? Whether the prime number comes in position 1, 2, 3 or 4, the possibility of getting one prime and three non-prime is the same, 1/16 regardless of the order, no? Same with probability of no primes, 1/16, regardless of the order.

So wouldn't the probability of having at least two primes be 1-(1/16+1/16)=1-2/16=1-1/8=7/8?

You have to multiply by 4 \(\frac{1}{2}^4\) in order to consider all the combinations.

P,N,N,N - N,P,N,N - N,N,P,N - N,N,N,P - each of those has a probability of \(\frac{1}{2}^4\) as you said, but there are 4 combinations in which we can find only one prime. That`s why you multiply by 4.

"Whether the prime number comes in position 1, 2, 3 or 4, the possibility of getting one prime and three non-prime is the same, 1/16 regardless of the order, no? Same with probability of no primes, 1/16, regardless of the order."

So, the answer to your question is : the order is important.

You don`t multiply the probability of no primes by 4, because there is only one way to get it N,N,N,N!

Hope its clear

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