516hw3 - STAT516 Solution to Homework 3 Section 2.1 4 1 6 1...

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STAT516 Solution to Homework 3 Section 2.1 2. 2 2 4 1 1 6 (2 boys & 2 girls) 0.375 2 2 2 16 P �� � �� � = = = ��� �� � � �� � �� Hence, (different number of boys & girls in a family of 4 children) 1 (2 boys & 2 girls) 1 0.375 0.625 P P = - = - = So, in a family of 4 children, different number of boys and girls would be more common than same number of boys and girls. And the relative frequencies are: (different number of boys & girls in a family of 4 children) 0.625 P = (same number of boys & girls in a family of 4 children) 0.375 P = 4. (2 sixes in first 5 rolls | 3 sixes in 8) P (2 sixes in first 5 3 sixes in 8) (3 sixes in 8) (2 sixes in first 5 1 six in last 3) (3 sixes in 8) (2 sixes in first 5) (1 sixes in last 3) [as first 5 and last 3 roll (3 sixes in 8) P P P P P P P i = = = 2 3 1 2 3 5 s are independent] 5 3 1 5 1 5 2 6 6 1 6 6 8 1 5 3 6 6 5 3 2 1 30 0.536 8 56 3 �� �� � �� � � �� � � �� � � �� � �� �� � �� � � �� � �� �� = �� � �� � � �� � �� � �� � �� ���� ���� ���� = = = �� �� ��
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6. a) ( 29 ( 29 4 4 8 (4 hits) 0.7 0.3 0.1361
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