Answer:
There are 10 counters in the bag and you want the 4
As their is only one for the probability is
1/10
There are now 9 counters left in the bag.
You only want the even ones , which are 2 6 8 and 10 ( four has been taken out)
There are only four even counters in the bag of nine counters so the probability is
4/9
As you want both of these to occur, you need to multiply them
1/10 x 4/9 = 49/90
Step-by-step explanation:
Sample space = {p, r, o, b, a, b, i, l, I, t, y} = 11 possible outcomes
1sr event: drawing an I ( there are 2 I); P(1st I) = 2/11
2nd event drawing also an i: This is a conditional probability, since one I has already been selected the remaining number of I is now 1, but also the sample space from previously 11 outcome has now 10 outcomes (one letter selected and not replaced)
2nd event : P(also one I) = 1/10
P(selecting one I AND another I) is 2/11 x 1/10
P(selecting one I AND another I) =2/110 = 0.018
<span>The orthocenter of a triangle is formed by the ALTITUDES of a triangle.</span>
Source:
http://www.1728.org/trictr.htm
Answer:
b
Step-by-step explanation: