To do this we need to move 10 to other side. To accomplish this you just need to add 10 to both side since (-10)
so
A+ 10 = c -10 + 10
we get
A+ 10 = c
lets say it wasn't -10 but positive 10.
A = c + 10 then we would subtract 10 from both sides
A -10 = c + 10 - 10
we get
A - 10 = C
It was chord less sorry if wrong
Probability of first getting a 3 = 4/52 = 1/13
The p(getting >2) = probability of getting one of the remaining 3's or 40 other cards ( Im taking the aces to be low value) = 43/51
Required probability = 1/13 * 43/51 = 43/663
If aces are counted high it will be 1/13 * 47/51 = 47/531
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