Question

A bag contains 16 coins of which 2 coins are counterfeit with heads on the both sides. The rest are fair coins. One is selected at random from the bag and tossed. The probability of getting a head is : (a) \(\frac{3}{16}\) (b) \(\frac{13}{16}\) (c) \(\frac{9}{16}\) (d) \(\frac{7}{16}\)

Short answer

Answer: The probability of getting a head is \(\frac{9}{16}\).

Step-by-step solution

Determine the probabilities of getting heads when tossing each type of coin

The probabilities of getting a head when tossing each type of coin are: - Probability of getting a head when tossing a counterfeit coin: \(P(H|C) = 1\) (since both sides have heads) - Probability of getting a head when tossing a fair coin: \(P(H|F) = \frac{1}{2}\) (since there is an equal probability for heads and tails)

Use the formula for conditional probability to determine the overall probability of getting a head

Now we can use the conditional probability formula to determine the overall probability of getting a head: \(P(H) = P(H \cap C) + P(H \cap F) = P(H|C) \cdot P(C) + P(H|F) \cdot P(F)\) Plugging in the values from steps 1 and 2, we get: \(P(H) = 1 \cdot \frac{2}{16} + \frac{1}{2} \cdot \frac{14}{16} = \frac{2}{16} + \frac{7}{16} = \frac{9}{16}\) Thus, the probability of getting a head when a coin is selected at random from the bag and tossed is \(\boxed{\text{(c)}\ \frac{9}{16}}\).

Key concepts

Conditional Probability

Understanding conditional probability is important, especially when dealing with various scenarios containing different conditions that can affect the outcome. It's a measure of the probability of an event occurring given that another event has already occurred. This concept is essential in our exercise involving coins.

To calculate the overall probability of getting heads from a mixed bag of fair and counterfeit coins, we apply conditional probability. In practical terms, we ask: 'What's the chance of getting heads, knowing that we've picked either a counterfeit or fair coin?' By answering this for both scenarios, we get a clearer picture of the total probability.

For instance, the probability of getting heads from a counterfeit coin, labelled as P(H|C), is a certainty (P(H|C) = 1), since both sides are heads. Conversely, for a fair coin, given the equal chance of heads or tails, this probability is P(H|F) = 0.5. By using the conditional probability formula we can find the overall likelihood of flipping heads, as shown in the step-by-step solution.

Probability of Getting a Head

When discussing the probability of getting a head, we are essentially talking about the chances of flipping a single coin and it landing on the heads side up. This might seem straightforward when dealing with a fair coin, where there are two equally likely outcomes - heads or tails. Under these conditions, the probability of getting a head P(H) is 1/2.

However, in the exercise provided, the scenario is more complex due to the presence of counterfeit coins. These fake coins have heads on both sides, thus the probability of getting a head is 1 upon tossing one. The overall probability changes when we combine the fair and counterfeit coins and requires us to calculate a weighted probability to account for the mixture in the bag.

Fair and Counterfeit Coins

The notion of 'fair' and 'counterfeit' coins introduces a degree of complexity into probability exercises. A fair coin is defined as one with an equal chance of landing on heads or tails, which implies no bias and follows the classic model of an unbiased coin flip. In contrast, a counterfeit coin in this context is designed to deceive, having heads on both sides, making the outcome of a flip predictable.

In the solved exercise, to accurately determine the overall probability of getting a head, it's crucial to consider the composition of the coin bag: the mix of 14 fair coins and 2 counterfeit ones. The overall probability is a sum of the probabilities of selecting and flipping these two different types of coins: one with an intrinsic probability of 1/2 for heads (fair coin) and another with a probability of 1 (counterfeit coin). This mix influences the combined probability dramatically as demonstrated in the given solution.