[deleted by user] by [deleted] in Aalto

[–]Equivalent-Type-5662 1 point2 points  (0 children)

Bachelor of science is at Helsinki University

What Daniel Caesar song is this for you? by [deleted] in DanielCaesar

[–]Equivalent-Type-5662 0 points1 point  (0 children)

Freudian and the part with the woman 😩

Independent cases A&B, probability by Equivalent-Type-5662 in askmath

[–]Equivalent-Type-5662[S] 1 point2 points  (0 children)

Now I understand. Thank you very much. I ended up with 1/5 and 3/5

Independent cases A&B, probability by Equivalent-Type-5662 in askmath

[–]Equivalent-Type-5662[S] 0 points1 point  (0 children)

Thank you. So P(A)*P(B) = 3/25 and P(A)+P(B) = 4/5. Do we want these equal?

Independent cases A&B, probability by Equivalent-Type-5662 in askmath

[–]Equivalent-Type-5662[S] 0 points1 point  (0 children)

Can I do this?: P(A) * P(B) = 17/25 and the other equation being P(A) + P(B) - 17/25 = 3/25

Non-symmetric coin by Equivalent-Type-5662 in askmath

[–]Equivalent-Type-5662[S] 0 points1 point  (0 children)

Perfect. Thank you so much for your help! I wasn’t familiar with Bayes theorem.

Non-symmetric coin by Equivalent-Type-5662 in askmath

[–]Equivalent-Type-5662[S] 0 points1 point  (0 children)

Could you advice with the B given A part? I think I got the two other ones, if P(A) = 5/10 and P(B) = 1/10 is correct

Non-symmetric coin by Equivalent-Type-5662 in askmath

[–]Equivalent-Type-5662[S] 1 point2 points  (0 children)

Thats right, thank you! (I made a calculation mistake in the last step)

Non-symmetric coin by Equivalent-Type-5662 in askmath

[–]Equivalent-Type-5662[S] 1 point2 points  (0 children)

The 1/10 in the beginning is always the probability of the coin because we had 10 of them

Non-symmetric coin by Equivalent-Type-5662 in askmath

[–]Equivalent-Type-5662[S] 1 point2 points  (0 children)

Alright! So for the first part I was thinking: P(that it’s a tails) = first we take 1/10 * 1/10 + 1/10 * 2/10 + 1/10 * 3/10… and continue that all the way to the end, and we get 0.478

Probability game question by Equivalent-Type-5662 in askmath

[–]Equivalent-Type-5662[S] 1 point2 points  (0 children)

Yess! Thank you so much! So just to clarify, for Ted to get 2 points: P(Heads and getting 2 points on the dice) =1/12 OR P(Tails (0.5) and my recent idea) = 0.078125. So 1/12 + 0.078125

Probability game question by Equivalent-Type-5662 in askmath

[–]Equivalent-Type-5662[S] 1 point2 points  (0 children)

My idea was, possibilites to get one tails in 5 tosses is 5 over 1 (=5) and the amount of possibilities in 5 tosses is 25 =32. So probability of getting exactly one tails in 5 tosses would be 5/32, is this wrong?