An honest coin is tossed 4 times in a row. Find the probability of each of the following events.
​(a) E Subscript 1​: ​"The coin comes up heads exactly​ twice."
​(b) E Subscript 2​: ​"All 4 tosses come up the​ same."
​(c) E Subscript 3​: ​"Exactly one-third of the tosses come up tails​."
​(d) E Subscript 4​: ​"The first 2 tosses come up tails​.

Let's solve the problem by breaking down each part.

We are tossing an honest (fair) coin 4 times. The outcomes of each toss are independent, and each toss has two possible outcomes: heads (H) or tails (T). The total number of possible outcomes of 4 tosses is \( 2^4 = 16 \).

### (a) \( E_1 \): "The coin comes up heads exactly twice."
To find the probability of getting exactly 2 heads in 4 tosses, we can use the binomial distribution formula. The number of ways to choose exactly 2 heads out of 4 tosses is given by the binomial coefficient:

\[
P(E_1) = \frac{\binom{4}{2}}{16}
\]

The binomial coefficient \( \binom{4}{2} \) represents the number of ways to choose 2 heads from 4 tosses:

\[
\binom{4}{2} = \frac{4!}{2!(4-2)!} = \frac{4 \times 3}{2 \times 1} = 6
\]

So, the probability is:

\[
P(E_1) = \frac{6}{16} = \frac{3}{8}
\]

### (b) \( E_2 \): "All 4 tosses come up the same."
For all 4 tosses to come up the same, either all 4 tosses must be heads (HHHH), or all 4 tosses must be tails (TTTT). These are the only two favorable outcomes out of 16 possible outcomes.

\[
P(E_2) = \frac{2}{16} = \frac{1}{8}
\]

### (c) \( E_3 \): "Exactly one-third of the tosses come up tails."
Since we are tossing the coin 4 times, it is impossible for exactly one-third of the tosses to be tails because one-third of 4 is not a whole number (it’s \( \frac{4}{3} \)). Therefore, this event is impossible.

\[
P(E_3) = 0
\]

### (d) \( E_4 \): "The first 2 tosses come up tails."
The first 2 tosses must be tails (TT), and the last 2 tosses can be anything (either heads or tails). There are 4 possible outcomes for the last 2 tosses: HH, HT, TH, TT. Thus, there are 4 favorable outcomes out of 16 possible outcomes.

\[
P(E_4) = \frac{4}{16} = \frac{1}{4}
\]

### Summary of probabilities:
- \( P(E_1) = \frac{3}{8} \)
- \( P(E_2) = \frac{1}{8} \)
- \( P(E_3) = 0 \)
- \( P(E_4) = \frac{1}{4} \)

Would you like further clarification on any part of this?

#### Related Questions:
1. What is the probability of getting at least one head in 4 coin tosses?
2. What is the probability of getting exactly 3 tails in 4 tosses?
3. How would the probabilities change if the coin were biased?
4. How many possible outcomes are there for tossing the coin 5 times?
5. What is the probability of getting exactly 1 head in 4 tosses?

**Tip:** Binomial coefficients are useful in determining the number of ways an event can happen in cases with independent trials like coin tosses.

An honest coin is tossed 4 times in a row. Find the probability of each of the following events. (a) E1: 'The coin comes up heads exactly twice.' (b) E2: 'All 4 tosses come up the same.' (c) E3: 'Exactly one-third of the tosses come up tails.' (d) E4: 'The first 2 tosses come up tails.'

Solve a probability problem involving 4 coin tosses. Learn how to find the probability of getting exactly 2 heads, all tosses the same, and other events. Key concepts include binomial distribution and combinatorics. Ideal for students in Grades 9-12.

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