Sure, let's break down the problem step-by-step to understand how we get to the final result.
1. Understand the Problem:
- We have a deck of 26 cards.
- We need to calculate the probability of two sequential events:
1. Drawing a card labeled "B".
2. Replacing the card back into the deck.
3. Drawing a card labeled "I".
2. Calculate the Probability of Individual Events:
a) Probability of Drawing a "B" on the First Draw:
- The number of cards is 26.
- The probability of drawing any specific card from the deck is [tex]\(\frac{1}{26}\)[/tex].
- Therefore, the probability of drawing a "B" is [tex]\(\frac{1}{26}\)[/tex].
b) Probability of Drawing an "I" on the Second Draw:
- After replacing the first card, the deck still has 26 cards.
- The probability of drawing any specific card remains [tex]\(\frac{1}{26}\)[/tex].
- Therefore, the probability of drawing an "I" on the second draw is also [tex]\(\frac{1}{26}\)[/tex].
3. Calculate the Combined Probability:
- Since the drawing of the cards are independent events (due to replacement), the combined probability of drawing first a "B" and then an "I" is the product of the probabilities of the two events.
Formula:
[tex]\[ \text{Combined Probability} = \text{Probability of drawing "B"} \times \text{Probability of drawing "I"} \][/tex]
[tex]\[ \text{Combined Probability} = \frac{1}{26} \times \frac{1}{26} \][/tex]
[tex]\[ \text{Combined Probability} = \frac{1}{676} \][/tex]
As a decimal, this probability is approximately:
[tex]\[ 0.0014792899408284025 \][/tex]
So, the final results are:
- The probability of drawing a "B" is approximately [tex]\(0.038461538461538464\)[/tex].
- The probability of drawing an "I" is approximately [tex]\(0.038461538461538464\)[/tex].
- The combined probability of drawing first a "B" and then an "I" is approximately [tex]\(0.0014792899408284025\)[/tex].
Therefore, the correct probability associated with the scenario given in the question is about [tex]\(0.0014792899408284025\)[/tex].
