Box a contains two red cards and one black card, and box b contains one red card and two black cards. an experiment consists of randomly selecting one card from box a and placing the selected card into box b. then, one card is selected from box b. what is the probability that a red card is selected from box b?

5/12

Step-by-step explanation:

Box A: 2 Red

           1 Black

Box B: 1 Red

           2 Black

One card is selected from Box A and placing the selected card into Box B.

One card is selected from Box B.

What is the probability that a red card is selected from Box B?

Two things can occur:

1) Select red from A and B:

2/3 * 2/4 = 1/3

2) Select black from A and red from B:

1/3 * 1/4 = 1/12

As 1) or 2) can occur:

1/3 + 1/12 = 5/12

< xcy ≅ < xby

step-by-step explanation:

in δcax and δbax,

so, δcax ≅ δbax (side-angle-side or sas)

since δcax ≅ δbax, we can conclude < acx ≅ < abx

in δacy and δaby,

so, δacy ≅ δaby (side-angle-side or sas)

since δacy ≅ δaby, we can conclude < acy ≅ < aby

now,

< acy ≅ < aby

=> < acx + < xcy ≅ < abx + < xby

=> < acx + < xcy ≅ < acx + < xby   (since < acx ≅ < abx already proved above)

subtracting < acx from both sides, we get

< acx + < xcy -< acx ≅ < acx + < xby -< acx

cancelling out < acx and -< acx from both sides, we get

< xcy ≅ < xby (proved)

answer: your answer is a