Three coins whose faces are marked 1 and 2 are tossed. The expected sum of numbers on their faces is
A
4
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B
4.5
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C
5
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D
6
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Solution
The correct option is A4.5 Probability of coin when it shows 1 =P(1)=12 Probability of coin when it shows 2 =P(2)=12 Expected sum of numbers for three coins is =1×P(1)+2×P(2)+1×P(1)+2×P(2)+1×P(1)+2×P(2) =(1×12+2×12)+(1×12+2×12)+(1×12+2×12) =4.5