Find the probability of winning second prize (that is, picking five of the six winning numbers) with a 6/44 lottery, as played in Connecticut, Missouri, Oregon, and Virginia. (Round the answer to five decimal places.)

The answer to the question is

The probability of winning second prize (that is, picking five of the six winning numbers) with a 6/44 lottery, as played in Connecticut, Missouri, Oregon, and Virginia is 3.49905×10⁻⁵≡ 0.00003 to five decimal places.

Step-by-step explanation:

The probability of winning the second prize or picking five of the six winning numbers) with a 6/44 lottery is given by

Number of 5 sets of numbers in 44 = ₄₄C₅ = 1086008 ways

Number of 5 set of winning numbers in 44 = 1

Number of ways of picking the last number to make it 6 numbers is given by

44 - 5 lucky numbers - The 1 winning number = 38

Therefore, there are 38 ways from 1086008 of selecting the 5 second place winning numbers

Therefore the probability of picking the 5 second place winning numbers is = 3.49905×10⁻⁵

27

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)