MHT CET · Maths · Probability
In a single throw of three dice, the probability of getting a sum at least 5 is
- A \(\frac{53}{54}\)
- B \(\frac{51}{54}\)
- C \(\frac{1}{54}\)
- D \(\frac{2}{3}\)
Answer & Solution
Correct Answer
(A) \(\frac{53}{54}\)
Step-by-step Solution
Detailed explanation
Here \(n(S)=6 \times 6 \times 6=216\)
Sum less than \(5 \equiv\{(1,1,1),(1,1,2),(1,2,1),(2,1,1)\}\)
Here \(\mathrm{P}(\) sum less than 5\()=\frac{4}{216}=\frac{1}{54}\)
\(\therefore \mathrm{P}(\) at least 5\()=\mathrm{P}(\geq 5)=1-\mathrm{P}( < 5)\)
\(=1-\frac{1}{54}=\frac{53}{54}\)
Sum less than \(5 \equiv\{(1,1,1),(1,1,2),(1,2,1),(2,1,1)\}\)
Here \(\mathrm{P}(\) sum less than 5\()=\frac{4}{216}=\frac{1}{54}\)
\(\therefore \mathrm{P}(\) at least 5\()=\mathrm{P}(\geq 5)=1-\mathrm{P}( < 5)\)
\(=1-\frac{1}{54}=\frac{53}{54}\)
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