$5$ coins tossed whose marks $2$ and $3$,then the probability of total sum $12$ is :
- $\dfrac{1}{{32}}$
- $\dfrac{1}{{16}}$
- $\dfrac{3}{{16}}$
- $\dfrac{5}{{16}}$
With 5 coins having faces 2 and 3, the total sum can range from 10 to 15. The number of ways to get a sum of 12 is equivalent to choosing how many coins show 3. If x coins show 3 and (5-x) coins show 2, 3x + 2(5-x) = 12, so x + 10 = 12, x = 2. The number of ways to choose 2 coins out of 5 is 5C2 = 10. Total outcomes are 2^5 = 32. Probability = 10/32 = 5/16.
To achieve a total sum of 12 when tossing 5 coins marked 2 and 3, there must be exactly three coins showing 3 and two coins showing 2, because (3 coins times 3) plus (2 coins times 2) equals 13, while exactly two coins showing 3 and three coins showing 2 yields a sum of 12. The number of ways to choose exactly two coins showing 3 from 5 tosses is calculated using combinations as 5C2. The probability is 5C2 divided by 2^5, which equals 10/32, and this simplifies to 5/16.