Question
A die is thrown once. Find the probability of getting
- a prime number;
- a number lying between 2 and 6
- an odd number.
✓
Answer
Number of all possible outcome (1, 2, 3, 4, 5, 6) = 6
- Let E be the event of getting a prime number. Then, the outcomes favourable to E are 2, 3 and 5.
Therefore, the number outcomes favourable to E is 3.
So, $\mathrm { P } ( \mathrm { E } ) = \frac { \text { Number of outcomes favourable to } \mathrm { E } } { \text { Number of all possible outcomes } } = \frac { 3 } { 6 } = \frac { 1 } { 2 }$ - Let E be the event of getting a number lying between 2 and 6.
Then, the outcomes favourable to E are 3, 4, and 5. Therefore, the number of outcomes favourable to E is 3.
So, $\mathrm { P } ( \mathrm { E } ) = \frac { \mathrm { Number~of~outcomes } \text { favourable to } \mathrm { E } } { \text { Number of all possible outcomes } } = \frac { 3 } { 6 } = \frac { 1 } { 2 }$ - Let E be the event getting an odd number.
Then, the outcomes favourable to E are 1, 3, and 5. Therefore, the number of outcomes favourable to E is 3.
So, $\mathrm { P } ( \mathrm { E } ) = \frac { \text { Number of outcomes favourable to } \mathrm { E } } { \text { Number of all possible outcomes } } = \frac { 3 } { 6 } = \frac { 1 } { 2 }$
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