We can use BT with the invading UFOs. What’s the chance that an abducting UFO is from Zargon?
(What is the chance that a UFO is from Zargon given that it abducts?)
P(Zargon|Abduct) = P(Abduct|Zargon) x P(Zargon)
P(Abduct)
P(Zargon|Abduct) =
3
/8 x
8
/20 = 24/160 = 1
/2
6
/20 6
/20
This is the answer we’ve already seen. iv. -AT LEAST ONEWe flip three coins. What is the chance that any (at least one) of them lands Heads? We could solve this problem the way we’ve already done: count the outcomes we’re looking for, and divide by the total possible outcomes. With three coins, A, B, and C, we have 23 = 8 possible outcomes.
{AT, BT, CT} {AH, BT, CT} {AT, BH, CT} {AT, BT, CH}
{AH, BH, CT} {AT, BH, CH} {AH, BT, CH} {AH, BH, CH}
Seven of the eight …show more content…
However, the solution can never get higher than 100%. Nor can it ever reach 100%. That would mean absolute certainty, but we can always imagine that every coin lands Tails.
The way to get a number that gets larger and larger, but never reaches 1, is to subtract a smaller and smaller number from 1. We’ve already seen the rule P(A) = 1 – P(Not-A). We can use that again here.
P(AL1X) = 1 – P(Not-AL1X) “Not at least one X” means “not even one X” (“no X’s”).
Not-Heads = Tails; therefore:
P(Not-AL1Heads) = P(1stT-AND-2 ndT-AND-3 rdT) = 1
/2 x
1
/2 x
1
/2 = 1
/8
P(AL1Heads) = 1 – P(1 stT-AND-2 ndT-AND-3 rdT) = 1 –
1
/8 = 7
/8
This trick is very useful for problems with large numbers. Suppose we roll 10 dice.
What’s the chance of rolling any 3’s?. With one dice, the chance of not rolling 3 is 5
/6. With 10 dice, it is (5
/6)
10
. Again, we just subtract that number from 1.
P(AL1Roll3With10Dice) = 1 – (
5
/6)
10 = 1 –
9,765,625/60,466,176 ≈ 0.838
What if we roll 20 dice? The solution is the same, only with a different exponent.
P(AL1Roll3With20Dice) = 1 – (
5
/6)
20 = 1 –
95,367,431,640,625/3,656,158,440,062,976 ≈ 0.974
The probability is now up to about 97.4%. It will keep increasing like this with more dice: higher and higher, without ever reaching
(What is the chance that a UFO is from Zargon given that it abducts?)
P(Zargon|Abduct) = P(Abduct|Zargon) x P(Zargon)
P(Abduct)
P(Zargon|Abduct) =
3
/8 x
8
/20 = 24/160 = 1
/2
6
/20 6
/20
This is the answer we’ve already seen. iv. -AT LEAST ONEWe flip three coins. What is the chance that any (at least one) of them lands Heads? We could solve this problem the way we’ve already done: count the outcomes we’re looking for, and divide by the total possible outcomes. With three coins, A, B, and C, we have 23 = 8 possible outcomes.
{AT, BT, CT} {AH, BT, CT} {AT, BH, CT} {AT, BT, CH}
{AH, BH, CT} {AT, BH, CH} {AH, BT, CH} {AH, BH, CH}
Seven of the eight …show more content…
However, the solution can never get higher than 100%. Nor can it ever reach 100%. That would mean absolute certainty, but we can always imagine that every coin lands Tails.
The way to get a number that gets larger and larger, but never reaches 1, is to subtract a smaller and smaller number from 1. We’ve already seen the rule P(A) = 1 – P(Not-A). We can use that again here.
P(AL1X) = 1 – P(Not-AL1X) “Not at least one X” means “not even one X” (“no X’s”).
Not-Heads = Tails; therefore:
P(Not-AL1Heads) = P(1stT-AND-2 ndT-AND-3 rdT) = 1
/2 x
1
/2 x
1
/2 = 1
/8
P(AL1Heads) = 1 – P(1 stT-AND-2 ndT-AND-3 rdT) = 1 –
1
/8 = 7
/8
This trick is very useful for problems with large numbers. Suppose we roll 10 dice.
What’s the chance of rolling any 3’s?. With one dice, the chance of not rolling 3 is 5
/6. With 10 dice, it is (5
/6)
10
. Again, we just subtract that number from 1.
P(AL1Roll3With10Dice) = 1 – (
5
/6)
10 = 1 –
9,765,625/60,466,176 ≈ 0.838
What if we roll 20 dice? The solution is the same, only with a different exponent.
P(AL1Roll3With20Dice) = 1 – (
5
/6)
20 = 1 –
95,367,431,640,625/3,656,158,440,062,976 ≈ 0.974
The probability is now up to about 97.4%. It will keep increasing like this with more dice: higher and higher, without ever reaching