Answer:
a c e g
Step-by-step explanation:
yur welcume
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
Answer:
there are forty odd numbers
Step-by-step explanation:
Answer: yes your right
Step-by-step explanation: i used it it was right
