Binomial Probability Distributions | HSC Extension 1 Mathematics

[@TB-ob4ro]

when the maths is so boring that we don't pay attention in class the king comes and saves our atar.

[@emi-dv2fj]

For 2022 HSC question could you do this: Probability of selecting a green ball = 7/10 = 0.7 Hence, the expected number of green balls picked out of the 4 selections = 4 x 0.7 = 2.8 green balls Hence, the expected points from the green balls is 2.8 x (-5) = -14 points Probability of selecting a red ball = 3/10 = 0.3 Hence, the expected humber of red balls picked out of the 4 selections = 4 x 0.3 = 1.2 red balls Hence, the expected points from the red balls is 1.2 x 10 = 12 points Overall point score = 12-14 = -2 points

[@mcgrathematics]

@@emi-dv2fj that's really clever - I wish I'd done that, it's much simpler!

[@paqpaq4188]

Or, possibly even more simply you could just do: 3/10 x 10 + 7/10 x -5 (This is the expected score for just 1 trial) Then x4 (to simulate 4 trials) which = -2

[@TB-ob4ro]

that is exactly how i did it. alas i was an idiot and got the points mixed up ending up with an outcome of 22

[@DeapHere]

what about a miss*miss at 12:28?

[@mcgrathematics]

We're finding the probability of winning the game - which means we have to hit at least once 🙂

[@retroelkintg]

dont forget that z-scores can be calculated by mewing sigmas

[@jasmine_teaa]

You're a warrior for surviving the year 9's.. 💪💪

[@zongtaixue]

🐐

[@doobsta5871]

Cheers for the vid gunna save me for hsc

[@Josephxyz]

YOU DID IT

[@sheppy5152]

Thx