/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 110 A coin is tossed 3 times. Find t... [FREE SOLUTION] | 91影视

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Chapter 2: Problem 110

A coin is tossed 3 times. Find the probability that all 3 are heads, (a) if it is known that the first is heads, (b) if it is known that the first 2 are heads, (c) if it is known that 2 of them are heads.

Short Answer

Expert verified
The short version of the answer is: (a) P(A|B1) = \( \frac{1}{4} \) (b) P(A|B2) = \( \frac{1}{2} \) (c) P(A|B3) = 0

Step by step solution

01

Part (a): Probability of 3 heads given the first coin toss is heads

Let A represent the event that all 3 coin tosses are heads, and B1 represent the event that the first coin toss is heads. We want to find P(A|B1). We know that: 1. P(A) = P(3 heads) = (1/2) * (1/2) * (1/2) = 1/8 2. P(B1) = P(first toss is heads) = 1/2 Now we need to find P(A 鈭� B1), which is the probability of both A and B1 occurring. Since B1 is included in event A (all 3 coin tosses being heads), P(A 鈭� B1) is equal to P(A), which is 1/8. So, P(A|B1) = P(A 鈭� B1) / P(B1) = (1/8) / (1/2) = 1/4.
02

Part (b): Probability of 3 heads given the first 2 coin tosses are heads

Let B2 represent the event that the first 2 coin tosses are heads. We want to find P(A|B2). We know that: 1. P(A) = P(3 heads) = 1/8 2. P(B2) = P(first 2 tosses are heads) = (1/2) * (1/2) = 1/4 Now we need to find P(A 鈭� B2), which is the probability of both A and B2 occurring. Since B2 is included in event A (all 3 coin tosses being heads), P(A 鈭� B2) is equal to P(A), which is 1/8. So, P(A|B2) = P(A 鈭� B2) / P(B2) = (1/8) / (1/4) = 1/2.
03

Part (c): Probability of 3 heads given that 2 out of the 3 coin tosses are heads

Let B3 represent the event that 2 out of the 3 coin tosses are heads. We want to find P(A|B3). First, we will find P(B3). There are three possible ways to get 2 heads and 1 tail: HHT, HTH, and THH. Each of these has a probability of (1/2) * (1/2) * (1/2) = 1/8. So, P(B3) = 3 * (1/8) = 3/8. Now we need to find P(A 鈭� B3), which is the probability of both A and B3 occurring. Since A is the event of all 3 coin tosses being heads, it is not possible for it to occur together with B3 (2 out of the 3 coin tosses being heads). Hence, P(A 鈭� B3) = 0. So, P(A|B3) = P(A 鈭� B3) / P(B3) = 0 / (3/8) = 0.

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