Answer:
P=0.00564
Step-by-step explanation:
From Exercise we have 52 cards.
We calculate the number of combinations to draw 5 cards from a deck of 52 cards. We get
{52}_C_{5}=\frac{52!}{5!(52-5)!}=2598960
We now count the number of favorable combinations:
{13}_C_{1} · {48}_C_{2}= 13 · \frac{48!}{2!(48-2)!}=14664
Therefore, the probabilitiy is
14664/2598960=0.00564
P=0.00564
Answer:
1/4x9=9/4 2 1/4, 2.25, (3/2)2
Step-by-step explanation:
The answer is 9/4 when you multiply it. But it can also be simplified as one of the answers here I wrote above.
Answer:
B)113.04 cm^2
Step-by-step explanation:
I had this question before. The answer is 128 :)
Step-by-step explanation:
The data below is what was provided in the question and it is what I solved the question with
P(A1) = 0.23
P(A2) = 0.25
P(A3) = 0.29
P(A1 n A2 ) = 0.09
P(A1 n A3) = 0.11
P(A2 n A3) = 0.07
P(A1 n A2 n A3) = 0.02
a
P(A2|A1) = P(A1 n A2)/P(A1)
= 0.09/0.23
= 0.3913
We have 39.13% confidence that event A2 will occur given that event A1 already occured
b.)
P(A3 n A3|A1) = P(A2 n A3)n A1)/P(A1)
= 0.02/0.23
= 0.08695
We have about 8.7% chance of events A2 and A3 occuring given that A1 already occured.
C.
P(A2 u A3|A1)
= P(A1 n A2)u(A1 n A3)/P(A1)
= P( A1 n A2) + P(A1 n A3) - P(A1 n A2 n A3) / P(A1)
= (0.09+0.11-0.02)/0.23
= 0.18/0.23
= 0.7826
We have 78.26% chance of A2 or A3 happening given that A1 has already occured.
