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Toss an unbalanced coin 10 times. The probability of getting a head of the balanced coin is 0.5. Find the probabilities that you obtain using table and formula

(a) More than 5 heads

(b) Fewer than 8 heads

© between 2 to 8, inclusive, heads


We are asked to find probabilities related to the number of heads obtained when tossing an unbalanced coin 10 times. The probability \( p \) of getting a head is 0.5, and we need to use both a binomial table and a formula.

Let’s work through each part of the question using the binomial distribution formula and the table.

### 1. **General Binomial Formula**:

The binomial distribution formula is:

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

where:
- \( n \) is the number of trials (here \( n = 10 \)),
- \( k \) is the number of successful outcomes (number of heads),
- \( p = 0.5 \) is the probability of success (getting a head),
- \( 1 - p = 0.5 \) is the probability of failure (getting a tail).

### 2. **Binomial Table**:
You have provided a binomial table showing cumulative probabilities, where \( n = 10 \), and \( p \) ranges from 0.05 to 0.50. For this question, since \( p = 0.5 \), we will directly use the values for \( p = 0.5 \) from the table.

Now let's solve the parts:

---

### (a) **More than 5 heads (P(X > 5))**

This is the probability of getting more than 5 heads, i.e., \( P(X > 5) = P(6) + P(7) + P(8) + P(9) + P(10) \).

Using the binomial table for \( n = 10 \) and \( p = 0.5 \), we first find the cumulative probability \( P(X \leq 5) \), and subtract it from 1.

From the table:
- \( P(X \leq 5) \) for \( p = 0.5 \) is 0.6230.

Thus, the probability of more than 5 heads is:

\[
P(X > 5) = 1 - P(X \leq 5) = 1 - 0.6230 = 0.3770
\]

---

### (b) **Fewer than 8 heads (P(X < 8))**

This is the probability of getting fewer than 8 heads, i.e., \( P(X < 8) = P(0) + P(1) + \cdots + P(7) \).

From the table, we can directly use the cumulative probability \( P(X \leq 7) \).

From the table:
- \( P(X \leq 7) \) for \( p = 0.5 \) is 0.9453.

Thus:

\[
P(X < 8) = P(X \leq 7) = 0.9453
\]

---

### (c) **Between 2 to 8 heads, inclusive (P(2 \leq X \leq 8))**

This means the probability of getting between 2 and 8 heads, including both 2 and 8. 

We can calculate this using the cumulative probabilities for \( P(X \leq 8) \) and \( P(X < 2) = P(X \leq 1) \).

From the table:
- \( P(X \leq 8) \) for \( p = 0.5 \) is 0.9893.
- \( P(X \leq 1) \) for \( p = 0.5 \) is 0.0107.

Thus, the probability of getting between 2 and 8 heads is:

\[
P(2 \leq X \leq 8) = P(X \leq 8) - P(X \leq 1) = 0.9893 - 0.0107 = 0.9786
\]

---

### Final Answers:
(a) Probability of more than 5 heads \( P(X > 5) = 0.3770 \)

(b) Probability of fewer than 8 heads \( P(X < 8) = 0.9453 \)

(c) Probability of getting between 2 and 8 heads \( P(2 \leq X \leq 8) = 0.9786 \)

---

Would you like further details on the calculation steps, or to explore more examples related to the binomial distribution?

Here are some related questions:
1. What is the probability of getting exactly 5 heads?
2. How would the probabilities change if the probability of heads was 0.6?
3. How can we use the binomial formula instead of the table to solve these?
4. How does cumulative probability relate to individual probabilities?
5. How do you calculate the expected value and variance of a binomial distribution?

**Tip**: When calculating binomial probabilities, using cumulative probabilities from a table can save time compared to using the binomial formula for each individual value.

Toss an unbalanced coin 10 times. The probability of getting a head of the balanced coin is 0.5. Find the probabilities that you obtain using table and formula (a) More than 5 heads, (b) Fewer than 8 heads, (c) Between 2 to 8, inclusive, heads

Learn how to solve a binomial probability problem involving tossing an unbalanced coin 10 times, with the probability of heads being 0.5. Find the probabilities of obtaining more than 5 heads, fewer than 8 heads, and between 2 to 8 heads inclusive using both the binomial distribution formula and table.

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