A biased die is twice as likely to show an even number as an odd number. If such a die is thrown twice, find the probability distribution of the number of sixes. Also, find the mean of the distribution.
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Solution and Explanation
Step 1: Assign probabilities
Let \( P(3) = P(5) = p \), so \( P(2) = P(4) = P(6) = 2p \).
As the total probability is 1: \[ P(1) + P(2) + P(3) + P(4) + P(5) + P(6) = 1 \implies 9p = 1 \implies p = \frac{1}{9} \] Thus, \( P(6) = 2p = \frac{2}{9} \), and \( P(\text{Not getting six}) = 1 - P(6) = \frac{7}{9} \).
Step 2: Define the random variable \( X \)
Let \( X \) represent the number of sixes. The possible values of \( X \) are \( 0, 1, 2 \).
Step 3: Compute probabilities for \( X \)
\[ P(X = 0) = \left( \frac{7}{9} \right)^2 = \frac{49}{81}, \quad P(X = 1) = 2 \cdot \frac{2}{9} \cdot \frac{7}{9} = \frac{28}{81}, \quad P(X = 2) = \left( \frac{2}{9} \right)^2 = \frac{4}{81} \]
Step 4: Probability distribution of \( X \)
Step 5: Compute the mean of \( X \)
The mean is given by: \[ \mu = \sum_{i=1}^{3} X_i \cdot P(X_i) = 0 \cdot \frac{49}{81} + 1 \cdot \frac{28}{81} + 2 \cdot \frac{4}{81} = \frac{28}{81} + \frac{8}{81} = \frac{36}{81} = \frac{4}{9} \]
Step 6: Final result
The probability distribution of \( X \) is:
The mean of the distribution is \( \frac{4}{9} \).
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Self-study helps students to build confidence in learning. It boosts the self-esteem of the learners. Recent surveys suggested that close to 50% of learners were self-taught using internet resources and upskilled themselves.
A student may spend 1 hour to 6 hours in a day in upskilling self. The probability distribution of the number of hours spent by a student is given below:
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(iii)(b) Find \( P(1 < X < 6) \).Self-study helps students to build confidence in learning. It boosts the self-esteem of the learners. Recent surveys suggested that close to 50% learners were self-taught using internet resources and upskilled themselves. A student may spend 1 hour to 6 hours in a day upskilling self. The probability distribution of the number of hours spent by a student is given below:
\[ P(X = x) = \begin{cases} kx^2 & {for } x = 1, 2, 3, \\ 2kx & {for } x = 4, 5, 6, \\ 0 & {otherwise}. \end{cases} \]
Based on the above information, answer the following:
It is known that 20% of the students in a school have above 90% attendance and 80% of the students are irregular. Past year results show that 80% of students who have above 90% attendance and 20% of irregular students get “A” grade in their annual examination. At the end of a year, a student is chosen at random from the school and is found to have an “A” grade. What is the probability that the student is irregular?
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