3 Unit Maths HSC Exam Revision (1 Viewer)

[random-1006]

In a bag there are 5 yellow, 4 blue and 3 red marbles. Three marbles are drawn without replacement. Find the probability that:

a. the marbles are all blue
b. the marbles are all different colours
c. there is one yellow and two red marbles

 

[bored of sc]

In a bag there are 5 yellow, 4 blue and 3 red marbles. Three marbles are drawn without replacement. Find the probability that:

a. the marbles are all blue
b. the marbles are all different colours
c. there is one yellow and two red marbles

a. (4/12)*(3/11)*(2/10)

b. Number of ways in which marbles can be chosen: 3! = 6. So we need to consider 6 probabilities and add them together.

First selection*second selection*third selection

Y*B*R + B*R*Y + R*Y*B + R*B*Y + Y*R*B + B*Y*R =

(5/12)*(4/11)*(3/10) + (4/12)*(3/11)*(5/10) + (3/12)*(5/11)*(4/10) + (3/12)*(4/11)*(5/10) + (5/12)*(3/11)*(4/10) + (4/12)*(5/11)*(3/10)

c.
Yellow first: Y*R*R = (5/12)*(3/11)*(2/10) ---------------- (1)
Yellow second: R*Y*R = (3/12)*(5/11)*(2/10) -------------- (2)
Yellow third: R*R*Y = (3/12)*(2/11)*(5/10) ---------------- (3)

Add (1), (2) and (3) together to give you the probability.

 

[random-1006]

a. (4/12)*(3/11)*(2/10)

b. Number of ways in which marbles can be chosen: 3! = 6. So we need to consider 6 probabilities and add them together.

First selection*second selection*third selection

Y*B*R + B*R*Y + R*Y*B + R*B*Y + Y*R*B + B*Y*R =

(5/12)*(4/11)*(3/10) + (4/12)*(3/11)*(5/10) + (3/12)*(5/11)*(4/10) + (3/12)*(4/11)*(5/10) + (5/12)*(3/11)*(4/10) + (4/12)*(5/11)*(3/10)

c.
Yellow first: Y*R*R = (5/12)*(3/11)*(2/10) ---------------- (1)
Yellow second: R*Y*R = (3/12)*(5/11)*(2/10) -------------- (2)
Yellow third: R*R*Y = (3/12)*(2/11)*(5/10) ---------------- (3)

Add (1), (2) and (3) together to give you the probability.

theres a much quicker way, e.g. for part c

you figure out the probability of one of the outcomes, and then because you have 3 elements, two alike, they can be arranged 3! / 2! ways in a straight line ( ie 3 permutations)

so you do P ( YRR) * 3!/ 2!, its obvious from the three cases that you came up with are all the same value, you can apply the same sort of thing to part b as well

 

[random-1006]

A fair, six-sided die is thrown seven times. What is the probability that a ‘6’​

occurs on exactly 2 of the 7 throws?

 

[pOtentiaLeaver]

~binomial(2;7,1/6) ?

 

[random-1006]

~binomial(2;7,1/6) ?

yes, a reasonably well hidden binomial question, most of them are a bit more obvious

 

[deterministic]

well hidden????

gees i wonder what the more obvious ones look like

 

[random-1006]

well hidden????

gees i wonder what the more obvious ones look like

see last yrs hsc, usually the question is longer, and has more parts, much easier to spot

yes in the above question the "exactly" is a clue, but ive seen much more obvious questions