There are somureftobunters and some whiteeounters in a bag. At the start, $7$ of the counters are red and the rest of the counters are whitc. Woody takes two counters from the bag. First he takes at random a counter from the bag. He does not put the counter back in the bag. Woody then takes at random another counter from the bag. $\newline$ (a) Let the number of white counters in the bag be $x$ . $\newline$ Draw a tree diagram that represents the above scenario, showing all relevant information. counter Woody takes is red is $\frac{21}{80}$ . $\newline$ Find the number of white counters in the bag at the start. $\newline$ $2$

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Q. There are somureftobunters and some whiteeounters in a bag. At the start,

7

of the counters are red and the rest of the counters are whitc. Woody takes two counters from the bag. First he takes at random a counter from the bag. He does not put the counter back in the bag. Woody then takes at random another counter from the bag.

\newline

(a) Let the number of white counters in the bag be

x

\newline

Draw a tree diagram that represents the above scenario, showing all relevant information. counter Woody takes is red is

\frac{21}{80}

\newline

Find the number of white counters in the bag at the start.

\newline

2

Understand the Problem: First, let's understand the problem. We know that the probability of picking a red counter followed by another red counter is $\frac{21}{80}$ . We also know there are $7$ red counters and $x$ white counters. The total number of counters initially is $7 + x$ .
Calculate First Red Probability: Calculate the probability of picking the first red counter. This is $\frac{7}{7+x}$ because there are $7$ red counters out of $7 + x$ total counters.
Calculate Second Red Probability: After removing one red counter, there are $6$ red counters left and $x$ white counters, so the total is $6 + x$ . The probability of picking another red counter now is $\frac{6}{6+x}$ .
Calculate Joint Probability: The probability of both events (picking two red counters in a row) happening is the product of the probabilities of each event. So, $\frac{7}{7+x} \times \frac{6}{6+x} = \frac{42}{(7+x)(6+x)}$ .
Set Up Equation: Set up the equation based on the given probability: $\frac{42}{(7+x)(6+x)} = \frac{21}{80}$ .
Cross-Multiply: Cross-multiply to solve for $x$ : $42 \times 80 = 21 \times (7+x)(6+x)$ .
Simplify and Solve: Simplify and solve the equation: $3360 = 21 \times (42 + 13x + x^2)$ .
Rearrange Equation: Divide both sides by $21$ : $160 = 42 + 13x + x^2$ .
Use Quadratic Formula: Rearrange to form a quadratic equation: $x^2 + 13x - 118 = 0$ .
Calculate Discriminant: Use the quadratic formula to solve for $x$ : $x = \frac{-13 \pm \sqrt{169 + 472}}{2}$ .
Round to Nearest Whole Number: Calculate the discriminant and solve: $x = \frac{-13 \pm \sqrt{641}}{2}$ .
Round to Nearest Whole Number: Calculate the discriminant and solve: $x = \frac{-13 \pm \sqrt{641}}{2}$ . Since $x$ must be a whole number (you can't have a fraction of a counter), round $\sqrt{641}$ to the nearest whole number, which is approximately $25$ . $3$ . So, $x = \frac{-13 \pm 25.3}{2}$ .