Question

A fair die is thrown once. Find the probability of getting (a) a prime number (b) a multiple of 3

Answer 1

Answer:

Total outcomes in a die= 6 ( 1,2,3,4,5,6)

No. of possible outcomes = 6

Favourable outcomes i.e. a prime number= 2,3,5

Total favourable outcomes =3

Probability of getting a prime number

$= \frac{3}{6} \\ \\ = \frac{ \cancel3}{ { \cancel6} \: ^{2} } \\ \\ = \frac{1}{2}$

2. Total outcomes = 6

So there are 2 multiples of 3 i.e. 3 and 6

Probability =

$\frac{2}{6} \\ \\ = \frac{ \cancel2}{ { \cancel6} \: \: ^{3} } \\ \\ = \frac{1}{3}$

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Additional information:-

• Probability is the branch of mathematics concerning numerical descriptions of how likely an event is to occur, or how likely it is that a proposition is true. The probability of an event is a number between 0 and 1, where, roughly speaking, 0 indicates impossibility of the event and 1 indicates certainty.
• Probability. P(A) refers to the probability that event A will occur. P(A|B) refers to the conditional probability that event A occurs, given that event B has occurred. P(A') refers to the probability of the complement of event A.
• A probability is a number that reflects the chance or likelihood that a particular event will occur. A probability of 0 indicates that there is no chance that a particular event will occur, whereas a probability of 1 indicates that an event is certain to occur.

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Answer 2

Answer:

(a) a prime number

$= \frac{1}{2} .$

(b) a multiple of 3

$\frac{1}{3} .$