If one card is drawn from an ordinary deck of cards what is the probability of getting a red card or a 10 or a queen?

Step-by-step explanation:

In probability

"AND" means to multiply, and

"OR" means to add

Here, we want probability getting a red card OR a 10 OR a queen. So we find each individual probabilities and ADD.

In a standard deck, there is 52 cards.

P(Red):

half the cards are red and half are black, so there are 52/2 = 26 red cards

Hence

P(Red) = 26/52 = 1/2

P(10):

There are 4 suits with each one having 1 "10" card, so there are FOUR 10s in a deck. So,

P(10) = 4/52 = 1/13

P(Queen):

There are 4 suits with each one having 1 Queen. So in 4 suits, there are a total of 4 Queens. So,

P(Queen) = 4/52 = 1/13

P(Red, or, 10, or, Queen) = 1/2 + 1/13 + 1/13 =

hreryryrranswer:

step-by-step explanation:

< 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)