Stupid assignment that the teacher didn't even teach us!!!?

Question

A student keeps 10 pens in a drawer. Although the pens all look the same, 2 are red, 3 are black, and the rest are blue.

the student takes two pens from the drawer at random, ne after the other. Assuming that the pens are not replaced:

1. draw a probability tree to show the possible outcomes from two picks.
2. find the probability of picking both of the pens thathave red ink.
3. find the probability of picking at least one pen with black ink.

Answer 1

1.Let the red ones are R0, R1; black B2, B3, B4; blue: L5, L6, L7, L8, L9, then the part of the tree will look like:

|

|***|***|***|***|***|***|***|***|

R0 R1 B2 B3 B4 L5 L6 L7 L8 L9 /1st draw/

|

|***|***|***|***|***|***|***|***|

R1 B2 B3 B4 L5 L6 L7 L8 L9 . . . . . . . /2nd draw/

From R1 imagine a sub-tree with items R0 B2 B3 B4 L5 L6 L7 L8 L9 on the row, corresponding to the 2nd draw, then from B2 another similar one etc. finishing with a sub-tree with items R0 B2 B3 B4 L5 L6 L7 L8 from L9 above.

(I had some difficulties to write the above in text mode, but I suppose your imagination is good enough!)



2 It is 1/45 = 2/90. In the above tree you'll find 2 branches (R0, then R1 and R1, then R0) out of 90, or, alternatively you can find it as stated in other answers.



3. Paint the red and the blue items green. You'll have 3 black and 7 green items and the required probability is complementary to that both drawn not to be green, i.e.

1 - (7/10)*(6/9) = 1 - 14/30 = 16/30 = 8/15.

Answer 2

1) i don't really know how to draw it, sorry!





2) first off, the probability of picking one red ink is 2/10. After one pen is taken out, there are is 1 red pen left out of 9 pens. So the probability of pick the seocond red pen is 1/9



P = 2/10 * 1/9 = 1/45





3) At least 1 black in means:

Black and 1 different color + Black and Black



The question does not tell a specific color of the second pen, so the second pen can either be red or blue.



2 (3/10) * (7/9) + (3/10)*(2/9) = 8/15



EDIT:



yes, we can subtract the probability of picking no black ink from one, i completely agree with you on this. But, i think you made an error. it should be 1 - (7/10)(6/9) because the pens are NOT replaced. After picking 1 non-black ink, there are only 6 non-black inks left. So 1 - (7/10)(6/9) = 8/15

Answer 3

Lol. some instructors are like that. My seventh grade instructor took factors off using fact my t's have been too short. while you're that for the period of touch, ask your father and mom in case you will get a private instruct, or request a variety exchange. do no longer motel to wagging english. you additionally can no longer gel along with her variety of coaching. some human beings have diverse strategies of studying (seen spacial, logical and so on.). with a bit of luck you will get a sparkling instructor next year. :) do no longer provide her any reason to care approximately your type. My sose instructor would have had 6 months off whilst she crashed her motor vehicle, yet she cared approximately my type too a lot and got here back day after right this moment, stressing approximately how a lot she cared for my type.

Answer 4

1) cant post in text-format

2) (2/10) * (1/9) = 2.2%

3) 100% - (7/10)*(7/9) = 45.6%



re. Rec: the sum of the probabilities will not yield the correct result. The "at least 1 instance" problem is calculated by taking 100%, and then subtracting the probability that there will be NO instances of the occurrence.

Answer 5

1. 7/10 for black pens, 1/5 for red pens and 3/10 for blue pens.

2. 1/5 for two picks for red pens.

3. 3/10 for black pens.

I think theses answers could be right...

Answer 6

The assignment is easier than you think.



IE: When you role a dice (di), what are your possible outcomes? 1-6



Possibilities for first role: 1-6

Possibilities for even number: 3/6

Possibilities for odd number: 3/6



etc...