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joyce
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prude_sb
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There is a formula for calucating probabilty for repeating events

See attached doc
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SDOC3320.pdf [28.24 KiB]
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ywilfred
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We need at least 3 heads, so the words that we need to arrange are:

HHHTT
HHHHT
HHHHH

Each case has a probability of 1/32

Case 1: HHHTT -> number of combinations = 5!/3!2! = 10. P = 10/32 = 5/16
Case 2: HHHHT -> number of combinations = 5!/4!1! = 5. P = 5/32
Case 3: HHHHH -> only 1 case -> 1/32

Total = 5/16 + 6/32 = 5/16 + 3/16 + 8/16 = 1/2

C
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andrehaui
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number of possible cases=10 as C5,3
number of total tentatives=5

so 10/5 = 1/2


is this way of solving correct for you?



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