Math Problem Statement

The probabilities that James and Juliet will pass an examination are 3/4 and 3/4 respectively. Find the probability that both will fail the examination. The table shows the distribution of balls in a bag. If 2 balls are selected at random with replacement, find the probability of selecting either 2 new green balls or 2 old blue balls.

Solution

Let's solve the problems in the image.

5a) The probability that James and Juliet will pass the examination are 34\frac{3}{4} and 34\frac{3}{4} respectively. Find the probability that both will fail the examination.

  • The probability of failing is the complement of the probability of passing.
  • For James, the probability of failing is 1−34=141 - \frac{3}{4} = \frac{1}{4}.
  • For Juliet, the probability of failing is also 1−34=141 - \frac{3}{4} = \frac{1}{4}.

Since the events are independent, the probability that both fail is the product of the probabilities:

Probability that both fail=14×14=116\text{Probability that both fail} = \frac{1}{4} \times \frac{1}{4} = \frac{1}{16}

So, the probability that both will fail the examination is 116\frac{1}{16}.

5b) The table shows the distribution of balls in a bag. We are asked to find the probability of selecting either:

  1. Two new green balls, or
  2. Two old blue balls.

The table is:

\hline \text{Balls} & \text{Green} & \text{Blue} \\ \hline \text{New} & 8 & 2 \\ \text{Old} & 4 & 6 \\ \hline \end{array}$$ The total number of balls is $$8 + 2 + 4 + 6 = 20$$. #### Case 1: Probability of selecting 2 new green balls - The number of new green balls is 8. - The probability of selecting a green ball on the first pick is $$\frac{8}{20}$$. - Since the selection is with replacement, the probability of selecting another green ball on the second pick is again $$\frac{8}{20}$$. The probability of selecting 2 new green balls is: $$\text{P(2 new green balls)} = \frac{8}{20} \times \frac{8}{20} = \frac{64}{400} = \frac{16}{100} = 0.16$$ #### Case 2: Probability of selecting 2 old blue balls - The number of old blue balls is 6. - The probability of selecting a blue ball on the first pick is $$\frac{6}{20}$$. - Since the selection is with replacement, the probability of selecting another blue ball on the second pick is again $$\frac{6}{20}$$. The probability of selecting 2 old blue balls is: $$\text{P(2 old blue balls)} = \frac{6}{20} \times \frac{6}{20} = \frac{36}{400} = \frac{9}{100} = 0.09$$ #### Total Probability Since we want either 2 new green balls or 2 old blue balls, we add the probabilities of these two events: $$\text{Total probability} = 0.16 + 0.09 = 0.25$$ Thus, the probability of selecting either 2 new green balls or 2 old blue balls is $$0.25$$. --- Would you like more details or have any further questions? Here are 5 related questions you might find useful: 1. How do we calculate the complement of a probability? 2. What changes in the calculation if the selection is without replacement? 3. How do we calculate the probability of multiple independent events? 4. How can we extend these concepts to three or more events? 5. What are the real-life applications of probability in decision making? **Tip**: When calculating probabilities for multiple events, always check if the events are independent or dependent, as this affects how probabilities are combined.

Ask a new question for Free

By Image

Drop file here or Click Here to upload

Math Problem Analysis

Mathematical Concepts

Probability
Complementary events
Independent events

Formulas

P(A') = 1 - P(A)
P(A and B) = P(A) * P(B)
P(event or event) = P(event1) + P(event2)

Theorems

Complement rule
Multiplication rule for independent events

Suitable Grade Level

Grades 9-10

Related Recommendation

Calculating Probability of Independent Events - Ammar and Azizah Exam Success

Probability of Independent Events A and B

Probability of Drawing Balls from a Box - Independent Events

Probability of Drawing Balls from a Container with and without Replacement

Probability of Drawing Even Numbers from a Bag with Replacement