Asked by Edward
From set of 20 natural no.s 2 are selected.
find probability that their sum is
1.odd
2.even
3.selected pair is twin prime
find probability that their sum is
1.odd
2.even
3.selected pair is twin prime
Answers
Answered by
Reiny
If we add two evens, we get an even
If we add two odds, we get an even
If we add an even and an odd, we get an odd
Let's look at this simplified scenario, of 4 numbers
+1 . 2 .3 .4
1: 2 3 4 5
2: 3 4 5 6
3: 4 5 6 7
4: 5 6 7 8
clearly half the sum's are even, half are odd
for 20 numbers:
prob(odd, odd)
= (10/20)(10/20) = 1/4
prob(even,even) = 1/4
both of these events yield an even sum
prob(even sum) = 1/4 + 1/4 = 1/2
clearly prob(odd sum) = 1- 1/2 = 1/2
"twin primes" are defined as two prime numbers that are two apart.
There are just a few from out set:
3,5 5,7 11,13 17,19 and their reverses
so we only have 10 such pairs
prob(twin primes) = 10/400 = 1/40
If we add two odds, we get an even
If we add an even and an odd, we get an odd
Let's look at this simplified scenario, of 4 numbers
+1 . 2 .3 .4
1: 2 3 4 5
2: 3 4 5 6
3: 4 5 6 7
4: 5 6 7 8
clearly half the sum's are even, half are odd
for 20 numbers:
prob(odd, odd)
= (10/20)(10/20) = 1/4
prob(even,even) = 1/4
both of these events yield an even sum
prob(even sum) = 1/4 + 1/4 = 1/2
clearly prob(odd sum) = 1- 1/2 = 1/2
"twin primes" are defined as two prime numbers that are two apart.
There are just a few from out set:
3,5 5,7 11,13 17,19 and their reverses
so we only have 10 such pairs
prob(twin primes) = 10/400 = 1/40
Answered by
Anonymous
Please tell me if no.s are selected randomly then what is
Answered by
Edward
If no.s are srlected randomly then what is probablity to above question
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