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Prove 3y² - 28y + 60 = 0 Using Conditional Probability of Drawing Socks
Mathematics
Grade 10 (Junior High School)
Question Content
Q4. DRAW A TREE DIAGRAM FOR THIS QUESTION – MAKE IT NEAT! There are y black socks and 5 white socks in a drawer. Joshua takes at random two socks from the drawer. (THIS MEANS ONE IS TAKEN, AND NOT REPLACED – i.e. conditional probability!) The probability that Joshua takes one white sock and one black sock is 6/11. (a) Show that 3y² - 28y + 60 = 0
Correct Answer
The equation 3y² - 28y + 60 = 0 is proven as required.
Detailed Solution Steps
1
Step 1: Calculate the total number of socks initially. There are y black socks and 5 white socks, so total socks = y + 5.
2
Step 2: Identify the two possible sequences for getting one white and one black sock: (white then black) or (black then white).
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Step 3: Calculate the probability of drawing white then black: Probability of first white sock is 5/(y+5). Since the sock is not replaced, the total number of socks left is y + 4, and the number of black socks is y, so the probability of second black sock is y/(y+4). The combined probability is (5/(y+5)) * (y/(y+4)).
4
Step 4: Calculate the probability of drawing black then white: Probability of first black sock is y/(y+5). After not replacing, total socks left is y + 4, number of white socks is 5, so probability of second white sock is 5/(y+4). The combined probability is (y/(y+5)) * (5/(y+4)).
5
Step 5: Add the two probabilities together, as they are mutually exclusive events, and set equal to 6/11: (5y/[(y+5)(y+4)]) + (5y/[(y+5)(y+4)]) = 6/11. This simplifies to 10y/[(y+5)(y+4)] = 6/11.
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Step 6: Cross-multiply to eliminate fractions: 11*10y = 6*(y+5)(y+4). Calculate both sides: 110y = 6(y² + 9y + 20).
7
Step 7: Expand the right-hand side: 110y = 6y² + 54y + 120.
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Step 8: Rearrange all terms to one side to form a quadratic equation: 0 = 6y² + 54y + 120 - 110y, which simplifies to 6y² - 56y + 120 = 0.
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Step 9: Divide the entire equation by 2 to simplify: 3y² - 28y + 60 = 0, which matches the equation we needed to prove.
Knowledge Points Involved
1
Conditional Probability (Without Replacement)
Conditional probability without replacement means that the outcome of the first event affects the probability of the second event, as the total number of items available decreases after the first selection. It is calculated by multiplying the probability of the first event by the probability of the second event given the first has occurred. This is used in scenarios like drawing objects from a set where items are not put back after selection.
2
Mutually Exclusive Events
Mutually exclusive events are events that cannot happen at the same time. For such events, the probability of either event occurring is the sum of their individual probabilities. In this problem, drawing white then black and black then white are mutually exclusive, so we add their probabilities.
3
Quadratic Equation Formation
Quadratic equations are polynomial equations of degree 2, in the form ax² + bx + c = 0. When solving probability problems involving unknown quantities, we can set up equations based on probability rules, then rearrange and simplify the equations to form quadratic equations for further solving.
4
Tree Diagrams for Probability
Tree diagrams are visual tools used to represent all possible outcomes of a sequence of events. Each branch represents an event, and the probability of each event is labeled on the branch. They help organize and calculate probabilities for multi-step events, especially conditional probability scenarios, by clearly showing the sequence and associated probabilities.
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