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In the context of our exercise, probability theory helps us determine the chance of picking a ball of a certain color from a box when the balls are selected with replacement. To calculate this, you would use the formula: \[\begin{equation} P(X) = \frac{\text{number of favorable outcomes}}{\text{total number of possible outcomes}}\end{equation}\]
Let A be the event that a head is tossed, and B be the event that an odd number is thrown. Directly from the sample space, calculate P(A∩B) and P(A ∪B). SOLUTION: (a) {Head,Tail} (b) {1,2,3,4,5,6} (c) {(1 ∩Head),(1 ∩Tail),...,(6 ∩Head),(6 ∩Tail)} Clearly P(A) = 1 2 = P(B). We can assume that the two events are independent, so
Solved probability problems and solutions are given here for a concept with clear understanding. Students can get a fair idea on the probability questions which are provided with the detailed step-by-step answers to every question.
For example, we might want to find the probability of drawing a particular 5-card poker hand. Since there are 52 cards in a deck and the order of cards doesn’t matter, the sample space for this experiment has 52C5 = 2,598, 52 C 5 = 2,598,960 possible 5-card hands.
Problem Solving - Basic. If I throw 2 standard 5-sided dice, what is the probability that the sum of their top faces equals to 10? Assume both throws are independent to each other. Solution: The only way to obtain a sum of 10 from two 5-sided dice is that both die shows 5 face up.
How should you distribute the balls so as to make the probability of choosing a red ball as large as possible and what will the probability be in that case? What happens if you have two bags, a hundred red balls and a hundred white balls?
Two balls are selected one by one without replacement. Find the probability that first is green and second is red. Sol: P (G) × P (R) = (5/12) x (7/11) = 35/132. Example 4: What is the probability of getting a sum of 7 when two dice are thrown? Sol: Probability math - Total number of ways = 6 × 6 = 36 ways.
