Probability of Drawing Same Letter
We need to find the probability that a letter drawn randomly from the word CAUSE is the same as a letter drawn randomly from the word EFFECT.
Letters in Each Word
Word 1: CAUSE
- Number of letters = 5
- Letters present: C, A, U, S, E (each appearing once)
Word 2: EFFECT
- Number of letters = 6
- Letters present: E (twice), F (twice), C (once), T (once)
Calculating Total Possible Outcomes
The total number of ways to draw one letter from CAUSE and one letter from EFFECT is the product of the number of letters in each word.
Total Outcomes = (Number of letters in CAUSE) $\times$ (Number of letters in EFFECT)
Total Outcomes = $5 \times 6 = 30$
Calculating Favorable Outcomes (Same Letter)
We need to find the pairs of letters that are the same from both words.
- Common Letters: The letters that appear in both words are 'C' and 'E'.
- Case 1: Drawing 'C'
- Ways to draw 'C' from CAUSE = 1
- Ways to draw 'C' from EFFECT = 1
- Number of pairs with 'C' = $1 \times 1 = 1$
- Case 2: Drawing 'E'
- Ways to draw 'E' from CAUSE = 1
- Ways to draw 'E' from EFFECT = 2
- Number of pairs with 'E' = $1 \times 2 = 2$
Total Favorable Outcomes = (Pairs with 'C') + (Pairs with 'E')
Total Favorable Outcomes = $1 + 2 = 3$
Determining the Probability
The probability is the ratio of favorable outcomes to total possible outcomes.
Probability = $\frac{\text{Total Favorable Outcomes}}{\text{Total Outcomes}}$
Probability = $\frac{3}{30} = \frac{1}{10}$
Important Questions from Probability (Notes)
- In a box there are 4 white balls and 6 black balls. A ball is drawn at random. If it is white, it is put back along with two more white balls in the box. If it is black, it is put back in the box and then two black balls are thrown out of the box. Now a ball is drawn again at random from the box. Then, what is the probability that it is black?
- A fair coin is tossed three times. Let A be the event of getting exactly two heads and B be the event of getting at most
two tails, then P(A$\cup$B) is: Bag A contains 3 Red and 4 Black balls while Bag B contains 5 Red and 6 Black balls. One ball is drawn at random from one of the bags and is found to be red. Then, the probability that it was drawn from Bag B is
- If we twice flip a balanced coin, what is the probability of getting at least one head?
1. 1/4
2. 2/4
3. 1/6
4. 3/4 Suppose that the random variable X takes on the values: -1, 0, and 2 with probability $\frac{1}{8}$, $\frac{1}{2}$ and $\frac{3}{8}$. Find the expected value of X.
