You are out with friends. half of you want to go bowling and the other half want to go to a movie. how will you make a fair decision about whether to go to a movie or go bowling using a fair coin and assuming that all of you want to go to either of the places together? (let h = heads and t = tails)

1) flip the coin twice. if the outcome is in the sequence hh, go to the movie. if it is not hh, go bowling.

2) flip the coin twice. if the outcome is in the sequence ht, go to the movie. if it is th, go bowling or repeat the process.

3) flip the coin three times. if the outcome is in the sequence hht, go to the movie. if it is ttt or hhh, go bowling; otherwise, repeat the process.

4) flip the coin three times. if the outcome is in the sequence hhh or ttt, go to the movie; otherwise, go bowling.

Option B

Step-by-step explanation:

Given that  Half of you want to go bowling and the other half want to go to a movie

So we must get probability for bowling = prob for movie

Option 1:  Here P(HH) = 1/4 and P(no HH) = 3/4

SInce the two are not equal this is not fair

2) Here P(HT ) =1/4 and P(TH)=1/4 and other outcomes repeat the process

This is fine and hence this option is right

3) P(HHT) = 1/8 and P(TTT or HHH) =1/8+1/8=1/4

This is biased and hence not fair

4) P(HHH or TTT) = 1/4 and P(Other options)=3/4

Biased and hence not fair

Only option B is right.

B

Step-by-step explanation:

took it on plato

let m = monomial

let t = trinomial

so, m times t = t.

monomial times a trinomial = another trinomial.

< 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)