PROBABILITY
- A round archery target of diameter 1 m is marked with four scoring regions from the centre outwards as red, blue, yellow and white. The radius of the red band is 0.20 m. The width of all the remaining bands is equal. If archers throw arrows towards the target, what is the probability that the arrows fall in the red region of the archery target?(2016)
(A) 0.40
(B) 0.20
(C) 0.16
(D) 0.04
Answer-C
To find the probability that the arrows fall in the red region of the archery target, we need to determine the ratio of the area of the red region to the total area of the target.
Given that the diameter of the target is 1 m, the radius of the red band is 0.20 m. This means the radius of the entire target is 0.50 m.
The area of a circle is calculated using the formula: A = πr^2, where A is the area and r is the radius.
The area of the red region can be calculated as follows:
A_red = π(0.20)^2 = 0.04π
The area of the entire target is:
A_total = π(0.50)^2 = 0.25π
Therefore, the probability of an arrow falling in the red region is given by:
P(red) = A_red / A_total = (0.04π) / (0.25π) = 0.04/0.25 = 0.16
Hence, the correct answer is (c) 0.16.
- 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?(2017)
(A) 17
(B) 16
(C) 13
(D) 11
Answer-B
To be assured of picking at least one ball of each color, we need to determine the worst-case scenario where we initially pick balls of the same color.
In this case, we need to find the minimum number of balls we must pick to ensure that we have picked all the balls of a particular color.
The worst-case scenario occurs when we first pick all the green balls, then all the white balls, and finally all the red balls.
Therefore, we need to pick the maximum number of balls of each color before we have picked at least one ball of each color.
The minimum number of balls to be picked can be calculated as follows:
Number of green balls = 8
Number of white balls = 7
Number of red balls = 5
We can guarantee that we have picked at least one ball of each color by picking all the green balls (8), all the white balls (7), and one more ball.
Therefore, the minimum number of balls to be picked is 8 + 7 + 1 = 16.
Hence, the correct answer is (b) 16.
- 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?(2018)
(A) 1/2
(B) 4/7
(C) 5/9
(D) 12/13
Answer-B
The total number of balls in the bag = 15 + 20 = 35
Let R be the event that a red ball is drawn and B be the event that a black ball is drawn.
Let 1, 2, and 3 be the events that a ball numbered 1, 2, or 3 is drawn, respectively.
P(R) = 15/35 = 3/7
P(B) = 20/35 = 4/7
P(3|R) = 40/100 * 3/7 = 6/35
P(1 or 2|B) = 1 - P(3|B) = 1 - 30/100 * 4/7 = 22/35
P(1 or 2 or 3|R or B) = P(1 or 2|R) * P(R) + P(1 or 2|B) * P(B)
= (1 - P(3|R)) * P(R) + (22/35) * (4/7)
= (1 - 6/35) * 3/7 + 22/35 * 4/7
= 129/245
Therefore, the chance of winning = P(1 or 2 or 3|R or B) = 129/245 ≈ 0.5265 ≈ 4/7
Hence, the answer is (b) 4/7.
