Question

Jim picked a card from a standard deck, looked at it, and then put it back. He then picked a second card. What is the probability that Jim picked a heart or an ace on either pick?

one over fifty two

two over fifty two

sixteen over fifty two

seventeen over fifty two

Answer 1

@Destinymasha

Answer 2

@Loser66 @arabpride

Answer 3

@sourwing @Compassionate

Answer 4

52 in a standard deck.

Four aces in a standard deck
Thirteen hearts in a standard deck

4/52 + 13/52 = 17/52

Answer 5

can you help me answer 2 more questions @Compassionate

Answer 6

Sorry, no... You caught me by luck... Try asking @Hero or @ranga

Answer 7

@ranga please help me with these last two questions!

Answer 8

@Kainui

Answer 9

i posted the questions as a snapshot!

Answer 10

\[\sum_{n=3}^{12}20(0.5)^{n-1} = 20\sum_{n=3}^{12}(0.5)^{n-1} = 20(0.5^2 + 0.5^3+....+0.5^{11})\]Use sum of geometric series formula to add up the numbers in the parenthesis.

Answer 11

Sum of geometric series:\[a + ar + ar^2 + .... + ar^{n-1} = a\frac{1-r^n}{1-r}\\ \text{Here }a = 0.5^2;~~r = 0.5;~~n = 10\]

Answer 12

I get 0.124755859

Answer 13

@ranga

Answer 14

That is not what I am getting.

Answer 15

\[20(0.5^2 + 0.5^3+....+0.5^{11}) = 20 * 0.5^2 * \left(\frac{1 - 0.5^{10}}{1 - 0.5}\right) = ?\]

Answer 16

When using calculator be sure to make use of parenthesis.

Answer 17

I got the answer! @ranga can you help me with the question below it!

Answer 18

What answer did you get?

Answer 19

9.99

Answer 20

Yes!

Answer 21

For the last one, they want a sigma notation to sum term 4 through term 15. That gives a hint the sigma notation ought to be \(\sum_{n=4}^{15}\) and so it has got to be either 1 or 3. Try n = 1 in 1 and 3 and see which one gives the first term as 5 (because in the table when n = 1, \(a_n = a_1 = 5\).

Answer 22

@ganeshie8

Answer 23

would the correct answer be C @ranga

Answer 24

@myininaya

Answer 25

@ganeshie8

Answer 26

First we eliminated the second and fourth choices because the sigma runs from n=1 to 15 in those choices whereas the problem states they want the sum from 4th term to 15th term and so the sigma should run from n = 4 to 15 which are the 1st and 3rd choices.

In the first choice, the nth term is indicated by the expression after the sigma which is: \(5(-2)^{n-1}\). If you put n = 1, you get the first term to be: \(5(-2)^{1-1}= 5(-2)^0 = 5*1 = 5\) which is what the table indicates the first term is when n = 1.

In the third choice, the nth term is: \(5(-2)^{n+1}\). If you put n = 1, you get the first term to be: \(5(-2)^{1+1}= 5(-2)^2 = 5*4 = 20\) which is NOT what the table indicates the first term is when n = 1.