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Cud anyone please answer question 6 part (I)
http://papers.xtremepapers.com/CIE/Cambridge International O Level/Statistics (4040)/4040_w10_qp_21.pdf
Ok Fine,
(i) For Throwing at first turn, Simply he got
0.2 of Probability
For Second, he Must Fail at First Turn so,
0.8 (Fail) x 0.2 (Hit) = 0.16
For Third, He Definatly must fail two times first.
So 0.8 (Fail) x 0.8 (Fail) x 0.2 (Hit) = 0.128
I Guess it Helps,
Best of Luck