From Randomness to Probability. HELP PLEASE

#1
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  • The table below gives partial data for 84 dogs cross-classified according to personality trait and breed. The cell counts for the Doberman and Retriever are missing, but row and column totals are given.
  • If one dog is randomly chosen from the 84, what is the chance that it is:
    i. Meek? 22/84 = 11/42 = 26.19%
    ii. An aggressive Rottweiler? 3/84 = 1/28 = 3.57%
    iii. Reactive or a Rottweiler? 60/84 = 5/7 = 71.43%
    iv. Neither aggressive nor a Rottweiler? 22/84 = 11/42 = 26.19%
    v. Meek if it is known that it is a Rottweiler? 2/84 = 1/42 = 2.38%
    vi. A Rottweiler if it is known to be aggressive? 3/84 = 1/28 = 3.57%
  • What value of x in the table makes the following two events independent? D = dog is a Doberman A = dog is Aggressive
    Support your answer with clear probability statements involving D and A.

    For the two events to be independent D and A, we must have P (A/D) = P (A)
    so… x/35 = 12/84
    x = 12/84 * 35
    x = 5