Answer:
![(3)/(16)](https://img.qammunity.org/2020/formulas/mathematics/college/hslisyy5lhryicphfmp3khxl6ad1eb0zlh.png)
Explanation:
Cards which are available are numbered as: {2, 3, 4, 5}
Number of cards available = 4
Selecting a number less than 3:
There is only one number less than 3, so only 1 way of selecting a card numbered less than 3. Therefore, the probability of selecting a card less than 3 =
![(1)/(3)](https://img.qammunity.org/2020/formulas/mathematics/middle-school/o08xg954t1gbzo9avralvfomcybk63rm02.png)
Since this card is placed back, the total number of cards in the sample remain the same.
Selecting a prime number:
There are 3 prime numbers in the sample: 2,3 and 5
Probability of selecting a prime numbers would be 3 out 4 i.e. =
![(3)/(4)](https://img.qammunity.org/2020/formulas/mathematics/college/e902xdvhskq5go8rzqwzc0u5ermj5vc0m5.png)
The two above mentioned events are independent of each other, so the probability of occurrence of both will be the product of their individual probabilities i.e.
Probability of drawing a number less than 3, putting the card back, then drawing a prime number =
![(1)/(4) * (3)/(4) = (3)/(16)](https://img.qammunity.org/2020/formulas/mathematics/middle-school/6px6ab8drvuswr5k62ed7y1n8024gqxyva.png)