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Out of 8 doctors, 6 arespecialists and 2 are MBBS. 3 doctors are selected in random. Find prob of distribution of the of the number of specialist doctors ?
There are three possibilities to select 3 people : (1) 1 is a specialist, 2 are non-specialists (X=1).(2) 2 are specialists, 1 is non-specialist (X=2).(3) all 3 are specialists (X=3).Also, not that X=0 is not a possibility as there can't be a case that all 3 selected are non-specialists, as there are only 2 non-specialists. So, the probability distribution of specialists is as follows:-P(X=0)=0P(X=1)=68×12×11=38P(X=2)=68×57×12=1556P(X=3)=68×57×46=514ThanksBharat BajajIIT Delhi
There are three possibilities to select 3 people :(1) 1 is a specialist, 2 are non-specialists (X=1).(2) 2 are specialists, 1 is non-specialist (X=2).(3) all 3 are specialists (X=3).Also, not that X=0 is not a possibility as there can't be a case that all 3 selected are non-specialists, as there are only 2 non-specialists. So, the probability distribution of specialists is as follows:-P(X=0)=0P(X=1)=68×12×11=38P(X=2)=68×57×12=1556P(X=3)=68×57×46=514ThanksBharat BajajIIT Delhi
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