Q 4. Maximum Shoe Attempts (Mathematics)
Raj has ten pairs of red, nine pairs of white and eight pairs of black shoes in a box. If he randomly picks shoes one by one (without replacement) from the box to get a red pair of shoes to wear, what is the maximum number of attempts he has to make?
2. Explanation:
To solve this problem, we need to determine the maximum number of attempts Raj might need to make to ensure he picks a pair of red shoes.
1. Understanding the Problem:
○ Raj has 10 pairs of red shoes, 9 pairs of white shoes, and 8 pairs of black shoes.
○ This means there are 10 \* 2 = 20 red shoes, 9 \* 2 = 18 white shoes, and 8 \* 2 = 16 black shoes in the box.
○ In total, there are 20 + 18 + 16 = 54 shoes in the box.
2. Objective:
○ Raj wants to pick a pair of red shoes. A pair means two shoes of the same color.
3. Worst-case Scenario:
○ To find the maximum number of attempts, consider the worst-case scenario where Raj picks as many non-red shoes as possible before picking a red pair.
○ He could pick all the white and black shoes first. This would be 18 (white) + 16 (black) = 34 shoes.
○ After picking all the non-red shoes, he would still need to pick two red shoes to form a pair.
4. Calculating Maximum Attempts:
○ After picking 34 non-red shoes, Raj will start picking red shoes.
○ To ensure he gets a pair, he might need to pick up to 2 more red shoes.
○ Therefore, the maximum number of attempts Raj might need is 34 (non-red shoes) + 2 (red shoes) = 36.
5. Re-evaluation:
○ However, the question asks for the maximum number of attempts to ensure a red pair, not just any pair.
○ In the worst-case scenario, Raj could pick all 44 non-red shoes (18 white + 16 black + 10 red singles) before finally picking the 45th shoe, which would complete a red pair.
Thus, the maximum number of attempts Raj might need to ensure he picks a pair of red shoes is 45.
Q 39. Probability
A bag contains 15 red balls and 20 black balls. Each ball is numbered either 1 or 2 or 3. 20% of the red balls are numbered 1 and 40% of them are numbered 3. Similarly, among the black balls, 45% are numbered 2 and 30% are numbered 3. A boy picks a ball at random. He wins if the ball is red and numbered 3 or if it is black and numbered 1 or 2. What are the chances of his winning?
- Total number of balls in the bag = 15 red + 20 black = 35
- Winning condition: If the ball is red and numbered 3 or if it is black and numbered 1 or 2
- Number of red balls numbered 3 = 40% of 15 = 6
- Number of black balls numbered 1 or 2 = 70% of 20 (25% foe number 1 + 45% for number 2)= 14
- Probability of winning = (6+14)/35 = 20/35 = 4/7
Q 66. Probability
A bag contains 20 balls. 8 balls are green, 7 are white and 5 are red. What is the minimum number of balls that must be picked up from the bag blindfolded (without replacing any of it) to be assured of picking at least one ball of each colour?
- A bag contains 20 balls. 8 balls are green, 7 are white and 5 are red. We have to pick at least one ball of each colour.
- Now consider this: Even if we pick 8 balls, there’s a probability that all of them may be green. Hence, we must pick more than 8 balls for sure.
- Now, even if we pick 8 + 7 balls, i.e. 15 balls, there’s a probability that all of them may be green and white only, with no red ball picked at all.
- Hence, taking the worst possible scenario in consideration we must pick 15 + 1 = 16 balls to be sure that atleast one ball of all the three colours is picked.
Q 25. Probability
Six identical cards are placed on a table. Each card has number '1' marked on one side and number '2' marked on its other side. All the six cards are placed in such a manner that the number '1' is on the upper side. In one try, exactly four (neither more nor less) cards are turned upside down. In how many least number of tries can the cards be turned upside down such that all the six cards show number "2' on the upper side?
2. Explanation:
To solve this problem, we need to determine the minimum number of tries required to have all six cards showing the number '2' on the upper side. Initially, all cards show '1'.
Step-by-step Process:
● Try 1: Turn any four cards upside down. Now, four cards show '2' and two cards show '1'.
● Try 2: Turn any four cards again. This can be done in such a way that the two cards showing '1' and two cards showing '2' are flipped. After this try, four cards will show '1' and two cards will show '2'.
● Try 3: Turn any four cards again. Choose the four cards that are currently showing '1'. After this try, all six cards will show '2'.
Therefore, the least number of tries required to achieve all cards showing '2' is 3.