MATH 又係probility

想問問part (a) 的答案先??
part (b) 的答案應該係 1 over ans. of part (a). 應該係 2點幾.
咁part (b) 答案就係 3. 因為要整數.

至於 P(a and b) = P(a)*P(b) , iff they are independent. 所以要事件係 indepentend先用到.

For example:

1. 第一次擲骰要係 6, 第二次都要係 6. 因為第一次擲骰後唔會影響第二次擲, 所以 independent.

2. 第一次擲骰要係 6, 第二次加埋第一次的總和要係 8, 因為第一次擲骰後會影響總和, 所以dependent.

3. If two cards are drawn with replacement from a deck of cards, the event of drawing a red card on the first trial and that of drawing a red card on the second trial are independent.

4. By contrast, if two cards are drawn without replacement from a deck of cards, the event of drawing a red card on the first trial and that of drawing a red card on the second trial are dependent.

Part A你識既我就唔慢慢講了，但B part無理由唔識架= = ? Nevermind , 我解釋一次‧

你宜家要每次抽一隻sock而唔會放番落個袋度 :
假如第一隻抽中黑色，而第二隻又抽中黑色 => p(B) x p(B) = 4/8 x 3/7 = 12/56
又假如第一隻抽中白色，而第二隻又抽中白色 => p(W) x p(W) = 4/8 x 3/7 = 12/56
由於12/56 + 12/56 = 24/56細過 1(概率=1即是100%機會中)，所以抽兩次未必一定抽得中兩隻都是同色‧

你試諗下，如果你抽頭兩隻都係唔同色，咁抽第三假一隻會同其中一隻色一樣啦~因為你得兩隻色(黑同白)，假如我先抽黑再抽白，第三隻無論抽到黑定白都有同色的兩隻socks

理論就係咁，至於用數學方式計都未嘗不可 :

B = 黑色，W =白色
B+W+B => p(B) x p(W) x p(B) = 4/8 x 4/7 x 3/6 = 1/7
B+W+W => p(B) x p(W) x p(W) = 4/8 x 4/7 x 3/6 = 1/7
W+B+B => p(W) x p(B) x p(B) = 4/8 x 4/7 x 3/6 = 1/7
W+B+W => p(W) x p(B) x p(W) = 4/8 x 4/7 x 3/6 = 1/7
W+W+任何色 => p(W) x p(W) x p(any) = 4/8 x 3/7 x 1 = 3/14
B+B+任何色 => p(B) x p(B) x p(any) = 4/8 x 3/7 x 1 = 3/14
點解要計埋W+W+any同B+B+any呢 ? 因為都係要抽三次，所以要計埋，之不過呢兩種抽法係第二次已經抽中‧
將以上既所有數加埋 : 1/7 x 4 + 3/14 x 2 = 7/7 即係 1
由於概率等於 1 ，所以抽三次係100%命中‧