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Your tree diagram is correct. However, you have listed 3 outcomes instead of 2. Adam and Betty choose drinks, so 2 outcomes. The third branch is not needed. For e.g.
P(adam choose orange and betty choose orange)=6/10 x 5/9=1/3
P(adam choose orange and betty choose orange)=6/10 x 5/9=1/3
P(win in first race and lose in 2nd)=0.8x0.2=0.16
P(win first 3races) =0.8 x 0.8 x0.8=0.512
P(takes part in race for n weeks)
= P(wins in every race for n weeks)
= 0.8^n
P(lose race in 1st week) or p(win in first week and lose in 2nd week)
=0.2+ 0.8x0.2
=0.2+0.16
=0.36
P(win first 3races) =0.8 x 0.8 x0.8=0.512
P(takes part in race for n weeks)
= P(wins in every race for n weeks)
= 0.8^n
P(lose race in 1st week) or p(win in first week and lose in 2nd week)
=0.2+ 0.8x0.2
=0.2+0.16
=0.36
done
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Date Posted:
7 months ago