Math Problem Statement

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​.

Solution

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 24=162^4 = 16.

(a) E1E_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(E1)=(42)16P(E_1) = \frac{\binom{4}{2}}{16}

The binomial coefficient (42)\binom{4}{2} represents the number of ways to choose 2 heads from 4 tosses:

(42)=4!2!(42)!=4×32×1=6\binom{4}{2} = \frac{4!}{2!(4-2)!} = \frac{4 \times 3}{2 \times 1} = 6

So, the probability is:

P(E1)=616=38P(E_1) = \frac{6}{16} = \frac{3}{8}

(b) E2E_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(E2)=216=18P(E_2) = \frac{2}{16} = \frac{1}{8}

(c) E3E_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 43\frac{4}{3}). Therefore, this event is impossible.

P(E3)=0P(E_3) = 0

(d) E4E_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(E4)=416=14P(E_4) = \frac{4}{16} = \frac{1}{4}

Summary of probabilities:

  • P(E1)=38P(E_1) = \frac{3}{8}
  • P(E2)=18P(E_2) = \frac{1}{8}
  • P(E3)=0P(E_3) = 0
  • P(E4)=14P(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.

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Math Problem Analysis

Mathematical Concepts

Probability
Binomial Distribution
Combinatorics

Formulas

Binomial coefficient: P(E1) = (4 choose 2) / 16
Basic probability formula: P(E) = favorable outcomes / total outcomes

Theorems

Binomial Theorem
Basic Probability Theorem

Suitable Grade Level

Grades 9-12