Proportionality>Making Predictions and Solutions Using Theoretical Probability(TEKS.Math.7.6.D)
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Texas 7th Grade Math › Proportionality>Making Predictions and Solutions Using Theoretical Probability(TEKS.Math.7.6.D)
A bag has 4 blue, 3 red, and 5 green marbles. What is $P(\text{blue})$ on one draw? Structure: single draw from a bag; without replacement (single draw).
$4/7$
$1/3$
$3/12$
$5/12$
Explanation
Favorable outcomes: 4 blue. Total outcomes: $4+3+5=12$. So $P(\text{blue})=\frac{4}{12}=\frac{1}{3}$.
What is $P(\text{sum}=8)$ when rolling two fair six-sided dice? Structure: roll two dice; independent (no replacement concept).
$1/6$
$6/36$
$4/36$
$5/36$
Explanation
There are $36$ equally likely ordered outcomes. Sums of $8$: $(2,6),(3,5),(4,4),(5,3),(6,2)$, so 5 favorable. $P=\frac{5}{36}$. Distractors use wrong totals or miscount favorable outcomes.
Flip a fair coin and roll a fair number cube. What is $P(\text{heads and even})$? Structure: coin + number cube; independent.
$1/4$
$3/6$
$3/8$
$1/2$
Explanation
Use multiplication for independent events: $P(H)\times P(\text{even})=\frac{1}{2}\times\frac{3}{6}=\frac{3}{12}=\frac{1}{4}$. Distractors treat totals incorrectly or ignore one part.
A bag has 4 red and 6 blue marbles. Two marbles are drawn without replacement. What is $P(\text{red then red})$? Structure: two draws; without replacement (dependent).
$4/25$
$4/10$
$2/15$
$3/10$
Explanation
Multiplication rule with changing denominator: $P(R_1)\times P(R_2\mid R_1)=\frac{4}{10}\times\frac{3}{9}=\frac{12}{90}=\frac{2}{15}$. Distractors treat draws as independent or use wrong totals.
A bag has 2 blue, 3 red, and 5 green marbles. Two marbles are drawn with replacement. What is $P(\text{blue then red})$? Structure: two draws; with replacement (independent).
$1/15$
$3/50$
$1/2$
$3/10$
Explanation
With replacement the denominator stays $10$: $P(B)\times P(R)=\frac{2}{10}\times\frac{3}{10}=\frac{6}{100}=\frac{3}{50}$. Distractors treat it as dependent or add probabilities.