標題:
MATH probability一題!!!!!!!THXX
a fair dice is thrown and then a fair coin is tossed a number of times equal to the number shown on the dice.find the probrablity of obtaining no head given that the number on the dice is 4 or 5 THXXXXXXX 更新: wrong!!!!!!!!!!!! 答案是3/64 努力^^ 更新 2: ========== 下面的那位朋友@@OK的 thx 但係點解唔係 3/32? 我都係咁計到: {1/6(1/2)^4}/(1/6) + {1/6(1/2)^5}/1/6=3/32?? 為甚麼不對?
最佳解答:
When the die is 4 Probability of no head = (1/2)^4 = 1/16 When the die is 5 Probability of no head = (1/2)^5 = 1/32 As the probability of getting 4 or 5 is equal, the required probability = (1/2)(1/16 + 1/32) = 3/64 2011-10-19 23:36:39 補充： 因為已經給了骰子的數字是4或5﹐因此是1/2而不是1/6
其他解答:
........................................................................... 2011-10-27 20:36:57 補充： die is 4 = (1/2)^4 = 1/16 die is 5 = (1/2)^5 = 1/32 = (1/2)(1/16 + 1/32) = 3/64|||||**No headappears = tail appears**P(head)= P(tail) = 0.5 , P (1 or 2 or 3 or 4 or 5 or 6) = 1/6P(nohead|no. is 4) + P(no head|no. is 5)= P(tail|no. is 4) + P( tail|no. is 5)= P(tail∩no. is 4)/P(no.is 4) + P(tail∩no. is 5)/P(no. is 5)= ((0.5)^4x (1/6)) / (1/6) + ((0.5)^5 x (1/6)) / (1/6)=(0.5)^4+ (0.5)^5=0.09375
MATH probability一題!!!!!!!THXX
此文章來自奇摩知識+如有不便請留言告知
發問:a fair dice is thrown and then a fair coin is tossed a number of times equal to the number shown on the dice.find the probrablity of obtaining no head given that the number on the dice is 4 or 5 THXXXXXXX 更新: wrong!!!!!!!!!!!! 答案是3/64 努力^^ 更新 2: ========== 下面的那位朋友@@OK的 thx 但係點解唔係 3/32? 我都係咁計到: {1/6(1/2)^4}/(1/6) + {1/6(1/2)^5}/1/6=3/32?? 為甚麼不對?
最佳解答:
When the die is 4 Probability of no head = (1/2)^4 = 1/16 When the die is 5 Probability of no head = (1/2)^5 = 1/32 As the probability of getting 4 or 5 is equal, the required probability = (1/2)(1/16 + 1/32) = 3/64 2011-10-19 23:36:39 補充： 因為已經給了骰子的數字是4或5﹐因此是1/2而不是1/6
其他解答:
........................................................................... 2011-10-27 20:36:57 補充： die is 4 = (1/2)^4 = 1/16 die is 5 = (1/2)^5 = 1/32 = (1/2)(1/16 + 1/32) = 3/64|||||**No headappears = tail appears**P(head)= P(tail) = 0.5 , P (1 or 2 or 3 or 4 or 5 or 6) = 1/6P(nohead|no. is 4) + P(no head|no. is 5)= P(tail|no. is 4) + P( tail|no. is 5)= P(tail∩no. is 4)/P(no.is 4) + P(tail∩no. is 5)/P(no. is 5)= ((0.5)^4x (1/6)) / (1/6) + ((0.5)^5 x (1/6)) / (1/6)=(0.5)^4+ (0.5)^5=0.09375
文章標籤
全站熱搜
留言列表
