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Sagot :
Let's break down the solution step-by-step:
When three coins are tossed, the total number of possible outcomes is 8. This is because each coin has 2 possible outcomes (Heads or Tails), and so for three coins, the total possibilities are calculated as [tex]\(2^3 = 8\)[/tex].
Next, let's define our events:
- Event [tex]\( H \)[/tex]: All three coins show Heads. This event has only one favorable outcome which is (HHH).
- Event [tex]\( K \)[/tex]: At least one coin shows Heads. This event includes all outcomes except for (TTT). So, the favorable outcomes are (HHH, HHT, HTH, THH, HTT, THT, TTH). Thus, there are 7 favorable outcomes for event [tex]\( K \)[/tex].
Now, let's calculate the probabilities:
1. The probability [tex]\( P(H) \)[/tex] of all three coins being Heads:
[tex]$ P(H) = \frac{\text{Number of favorable outcomes for } H}{\text{Total number of possible outcomes}} = \frac{1}{8} $[/tex]
2. The probability [tex]\( P(K) \)[/tex] of getting at least one Head:
[tex]$ P(K) = \frac{\text{Number of favorable outcomes for } K}{\text{Total number of possible outcomes}} = \frac{7}{8} $[/tex]
3. The probability that the outcome is all Heads given that at least one coin shows a Head [tex]\( P(H \mid K) \)[/tex]:
To find this, we use the conditional probability formula:
[tex]$ P(H \mid K) = \frac{P(H \cap K)}{P(K)} $[/tex]
Since event [tex]\( H \)[/tex] (all Heads) is a subset of event [tex]\( K \)[/tex] (at least one Head), [tex]\( P(H \cap K) = P(H) \)[/tex]. Therefore:
[tex]$ P(H \mid K) = \frac{P(H)}{P(K)} = \frac{\frac{1}{8}}{\frac{7}{8}} = \frac{1}{7} $[/tex]
Given the numbers, we fill out the questions accordingly:
1. The expression [tex]\( P( H \cap K ) \)[/tex] is simply [tex]\( P(H) \)[/tex] because [tex]\( H \)[/tex] is a subset of [tex]\( K \)[/tex], so:
[tex]$ \frac{7}{8} \square P( H \cap K ) = \frac{1}{8} $[/tex]
2. The probability that the outcome is all heads given that at least one coin shows a head is:
[tex]$ \frac{1}{7} $[/tex]
3. The probability of getting at least one head [tex]\( P(K) \)[/tex] is:
[tex]$ \frac{1}{8} \square = \frac{7}{8} $[/tex]
When three coins are tossed, the total number of possible outcomes is 8. This is because each coin has 2 possible outcomes (Heads or Tails), and so for three coins, the total possibilities are calculated as [tex]\(2^3 = 8\)[/tex].
Next, let's define our events:
- Event [tex]\( H \)[/tex]: All three coins show Heads. This event has only one favorable outcome which is (HHH).
- Event [tex]\( K \)[/tex]: At least one coin shows Heads. This event includes all outcomes except for (TTT). So, the favorable outcomes are (HHH, HHT, HTH, THH, HTT, THT, TTH). Thus, there are 7 favorable outcomes for event [tex]\( K \)[/tex].
Now, let's calculate the probabilities:
1. The probability [tex]\( P(H) \)[/tex] of all three coins being Heads:
[tex]$ P(H) = \frac{\text{Number of favorable outcomes for } H}{\text{Total number of possible outcomes}} = \frac{1}{8} $[/tex]
2. The probability [tex]\( P(K) \)[/tex] of getting at least one Head:
[tex]$ P(K) = \frac{\text{Number of favorable outcomes for } K}{\text{Total number of possible outcomes}} = \frac{7}{8} $[/tex]
3. The probability that the outcome is all Heads given that at least one coin shows a Head [tex]\( P(H \mid K) \)[/tex]:
To find this, we use the conditional probability formula:
[tex]$ P(H \mid K) = \frac{P(H \cap K)}{P(K)} $[/tex]
Since event [tex]\( H \)[/tex] (all Heads) is a subset of event [tex]\( K \)[/tex] (at least one Head), [tex]\( P(H \cap K) = P(H) \)[/tex]. Therefore:
[tex]$ P(H \mid K) = \frac{P(H)}{P(K)} = \frac{\frac{1}{8}}{\frac{7}{8}} = \frac{1}{7} $[/tex]
Given the numbers, we fill out the questions accordingly:
1. The expression [tex]\( P( H \cap K ) \)[/tex] is simply [tex]\( P(H) \)[/tex] because [tex]\( H \)[/tex] is a subset of [tex]\( K \)[/tex], so:
[tex]$ \frac{7}{8} \square P( H \cap K ) = \frac{1}{8} $[/tex]
2. The probability that the outcome is all heads given that at least one coin shows a head is:
[tex]$ \frac{1}{7} $[/tex]
3. The probability of getting at least one head [tex]\( P(K) \)[/tex] is:
[tex]$ \frac{1}{8} \square = \frac{7}{8} $[/tex]
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