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Question Number 25559 by das47955@mail.com last updated on 11/Dec/17

(1) In a single throw of three dice,find the probability of getting a total of (i) 4 (ii) at most 4 (iii) atleast 4

$$\left(\mathrm{1}\right)\:\mathrm{In}\:\mathrm{a}\:\mathrm{single}\:\mathrm{throw}\:\mathrm{of}\:\mathrm{three}\: \\ $$$$\mathrm{dice},\mathrm{find}\:\mathrm{the}\:\mathrm{probability}\:\mathrm{of}\:\mathrm{getting} \\ $$$$\mathrm{a}\:\mathrm{total}\:\mathrm{of} \\ $$$$\left(\mathrm{i}\right)\:\mathrm{4} \\ $$$$\left(\mathrm{ii}\right)\:\mathrm{at}\:\mathrm{most}\:\mathrm{4} \\ $$$$\left(\mathrm{iii}\right)\:\mathrm{atleast}\:\mathrm{4} \\ $$

Commented by prakash jain last updated on 11/Dec/17

total number of outcomes=6×6×6=216 4: total number of favorable outcomes 1+2+1 1+1+2 2+1+1 3 probability of getting a 4=(3/(216)) (ii) at most 4 (3 and 4) 1+1+1=3 probability of getting at most 4=(4/(216)) (iii) at least 4. there is only 1 unfavorable outcome 1+1+1=3 probability=1−(1/(216))=((215)/(216))

$$\mathrm{total}\:\mathrm{number}\:\mathrm{of}\:\mathrm{outcomes}=\mathrm{6}×\mathrm{6}×\mathrm{6}=\mathrm{216} \\ $$$$\mathrm{4}:\:\mathrm{total}\:\mathrm{number}\:\mathrm{of}\:\mathrm{favorable}\:\mathrm{outcomes} \\ $$$$\mathrm{1}+\mathrm{2}+\mathrm{1} \\ $$$$\mathrm{1}+\mathrm{1}+\mathrm{2} \\ $$$$\mathrm{2}+\mathrm{1}+\mathrm{1} \\ $$$$\mathrm{3} \\ $$$$\mathrm{probability}\:\mathrm{of}\:\mathrm{getting}\:\mathrm{a}\:\mathrm{4}=\frac{\mathrm{3}}{\mathrm{216}} \\ $$$$\left(\mathrm{ii}\right)\:\mathrm{at}\:\mathrm{most}\:\mathrm{4}\:\left(\mathrm{3}\:\mathrm{and}\:\mathrm{4}\right) \\ $$$$\mathrm{1}+\mathrm{1}+\mathrm{1}=\mathrm{3} \\ $$$$\mathrm{probability}\:\mathrm{of}\:\mathrm{getting}\:\mathrm{at}\:\mathrm{most}\:\mathrm{4}=\frac{\mathrm{4}}{\mathrm{216}} \\ $$$$\left(\mathrm{iii}\right)\:\mathrm{at}\:\mathrm{least}\:\mathrm{4}. \\ $$$$\mathrm{there}\:\mathrm{is}\:\mathrm{only}\:\:\mathrm{1}\:\mathrm{unfavorable}\:\mathrm{outcome} \\ $$$$\mathrm{1}+\mathrm{1}+\mathrm{1}=\mathrm{3} \\ $$$$\mathrm{probability}=\mathrm{1}−\frac{\mathrm{1}}{\mathrm{216}}=\frac{\mathrm{215}}{\mathrm{216}} \\ $$

Commented by Rasheed.Sindhi last updated on 11/Dec/17

Salute to my math-guru!

$$\mathrm{Salute}\:\mathrm{to}\:\mathrm{my}\:\mathrm{math}-\mathrm{guru}! \\ $$

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