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    Question
    SLPaper 2

    A six-sided biased cube is weighted in such a way that the probability of obtaining a “six” is 710{\frac{7}{10}}107​.

    1.

    The cube is tossed five times. Find the probability of obtaining at most three "sixes".

    [3]
    Verified
    Solution

    recognition of binomial (M1)

    X ~ B(5, 0.7)

    attempt to find P (X ≤ 3) (M1)

    =0.472 (= 0.47178)*(A1)

    [3 marks]

    2.

    The cube is tossed five times. Find the probability of obtaining the third "six" on the fifth toss.

    [3]
    Verified
    Solution

    recognition of 2 sixes in 4 tosses (M1) P (3rd six on the 5th toss)=[(42)×(0.7)2×(0.3)2]×0.7(=0.2646×0.7){= \left[ {\left( {\begin{matrix} 4 \\ 2 \end{matrix}} \right) \times \left({0.7}\right)^2 \times \left({0.3}\right)^2} \right] \times 0.7 \left( =0.2646 \times 0.7 \right)}=[(42​)×(0.7)2×(0.3)2]×0.7(=0.2646×0.7) A1 = 0.185 (= 0.18522) A1 [3 marks]

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