Topic:C<span>onditional Probability
</span>For any two events A and B, where P(B) ≠ 0, P( A | B ) = P( A ∩ B ) / P( B ) = P( B | A) * P(A) / P(B)..i.e..{the probability of A given B is equal to the probability of A and B divided by the probability of B}
<span>For a set of events A1, A2, A3, ... , An,
</span>hence,
P(B)
<span>= P(B and A1) + P(B and A2) + ... + P(B and An) </span>
= P(B | A1) * P(A1) + P(B | A2) * P(A2) + ... + P(B | An) * P(An)
P(M | N)
<span>= P(N | M) * P(M) / P(N) </span>
<span>= P(N | M) * P(M) / (P(N | M) * P(M) + P(N | M') * (1 - P(M)))
</span><span>Substitute Values.... you get ?
</span>
We must find UNIQUE combinations because choosing a,b,c,d... is the same as d,c,b,a...etc. For this type of problem you use the "n choose k" formula...
n!/(k!(n-k)!), n=total number of choices available, k=number of choices made..
In this case:
20!/(10!(20-10)!)
20!/(10!*10!)
184756
The answer is 8/32 because 8 x 4 = 32
We know that
in a right triangle
Applying the Pythagoras Theorem
c²=a²+b²
in this problem
c=√87 yd
a=√23 yd
b=?
so
b²=c²-a²-----> b²=(√87)²-(√23)²----> b²=87-23----> b²=64----> b=8 yd
the answer is
8 yd
Answer:
Not a function
Step-by-step explanation:
If you use the line/ pencil test ***I think that's what it's called...
you put a pencil flat on the graph and if it touches more than one point, it is not a function. This only works with some graphs.But it works with this one.
