# Out of 7 lottery tickets 3 are prize-winning tickets. If someone buys 4 tickets what is the probability of winning exactly one prize?

From the Binomial distribution:

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The probability of winning exactly one prize when buying 4 tickets can be calculated using the binomial probability formula:

[ P(X=k) = \binom{n}{k} \cdot p^k \cdot (1-p)^{n-k} ]

Where:

- ( n ) is the number of trials (4 tickets in this case)
- ( k ) is the number of successes (exactly one prize in this case)
- ( p ) is the probability of success on a single trial (probability of winning a prize)

Given that there are 7 tickets with 3 prize-winning tickets, the probability of winning a prize on a single ticket is ( p = \frac{3}{7} ).

Plugging these values into the formula:

[ P(X=1) = \binom{4}{1} \cdot \left(\frac{3}{7}\right)^1 \cdot \left(1-\frac{3}{7}\right)^{4-1} ]

[ P(X=1) = 4 \cdot \frac{3}{7} \cdot \frac{4}{7}^3 ]

[ P(X=1) = 4 \cdot \frac{3}{7} \cdot \frac{64}{343} ]

[ P(X=1) = \frac{768}{2401} \approx 0.3203 ]

So, the probability of winning exactly one prize when buying 4 tickets is approximately 0.3203 or 32.03%.

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When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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