QN's answer to QN's Junior College 1 H1 Maths Singapore question.

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QN
Qn's answer
19 answers (A Helpful Person)
1st
I didn’t understand their method, is it wrong?
J
J
3 years ago
Erm my answer was for iii), not iv)
J
J
3 years ago
Here's my working for iv) :


P(at least 2 red socks out of 3 now)

= P(all 3 red)
+ P(2 red, 1 other colour)


= P(all 3 red)
+ P(1st red, 2nd not red, 3rd red)
+ P(1st red, 2nd red, 3rd not red)
+ P(1st not red, 2nd red, 3rd red)


= 10/25 x 9/24 x 8/23
+ 10/25 x 15/24 x 9/23
+ 10/25 x 9/24 x 15/23
+ 15/25 x 10/24 x 9/23

= (10x9x8 + 3x10x15x9)/(25 x 24 x 23)

= 4770/13800

= 477/1380

= 159/460
QN
QN
3 years ago
My bad I send the wrong part
QN
QN
3 years ago
Even for part(iv),
Shouldn’t it be
P(all 3 red)
+ P(1st red x 2nd red x 3rd pink)
+ P(1st red x 2nd red x 3rd blue)
+ P(1st pink x 2nd red x 3rd red)
+ P(1st blue x 2nd red x 3rd red)
QN
QN
3 years ago
At least 2, and there are different probabilities when I first pick red, blue or pink socks
J
J
3 years ago
Yes, but realise something : you've missed out these probabilities.

P(1st red, 2nd blue, 3rd red)
P(1st red, 2nd pink, 3rd red)
QN
QN
3 years ago
Ah thank you
J
J
3 years ago
What I did in my working was , instead of separately computing the cases with ① 2 reds 1 pink and ②2 reds 1 blue,

I choose to combine them together and do the various orders of 2 red, 1 non-red

(Non-red already covers both pink and blue)


You can try to do for the separate case and you should still get 159/460
J
J
3 years ago
Edited