Problem 19 - A Complete Resource Book in Mathematics for JEE Main

Question

An unbiased coin is tossed. If the result is a head, a pair of unbiased dice is rolled and the number obtained by adding the numbers on the two faces is noted. If the result is a tail, a card from a well shufled pack of eleven cards numbered \(2,3,4, \ldots, 12\) is picked and the number on the card is noted. The probability that the noted number is either 7 or 8 , is (A) \(\frac{193}{792}\) (B) \(\frac{164}{792}\) (C) \(\frac{231}{792}\) (D) none of these

Step-by-Step Solution

Answer

(A) \( \frac{193}{792} \).

Step 1: Define the Sample Space for the Coin Toss

The coin can either show a head or a tail. Therefore, the sample space for the coin toss is \( \{ H, T \} \), where \( H \) represents a head and \( T \) represents a tail.

Step 2: Probability for Head and Result from Dice

If the coin shows a head \( (H) \), we roll two dice. The sample space for rolling two dice is 6 \( \times \) 6 = 36 outcomes. We need to find the probability of obtaining a sum of either 7 or 8. The outcomes with sum 7 are \((1,6), (2,5), (3,4), (4,3), (5,2), (6,1)\), making 6 outcomes. The outcomes with sum 8 are \((2,6), (3,5), (4,4), (5,3), (6,2)\), making 5 outcomes. So the probability of getting 7 or 8 is \((6 + 5) / 36 = 11/36\).

Step 3: Probability for Tail and Picking a Card

If the coin shows a tail \( (T) \), you pick a card from 11 cards numbered 2 to 12. The numbers 7 and 8 each have 1 card available. Therefore, the probability of picking either 7 or 8 from the deck is \(2/11\).

Step 4: Calculate Overall Probability

Using total probability, the probability of the noted number being either 7 or 8 is the sum of probabilities for the head and tail case. \[P(7 \text{ or } 8) = P(T) \cdot P(7 \text{ or } 8 | T) + P(H) \cdot P(7 \text{ or } 8 | H)\]Here, \(P(T) = 1/2\) and \(P(H) = 1/2\). Substituting the values:\[P(7 \text{ or } 8) = \frac{1}{2} \cdot \frac{2}{11} + \frac{1}{2} \cdot \frac{11}{36} = \frac{1}{11} + \frac{11}{72}\]

Step 5: Simplify and Determine the Final Probability

We find a common denominator for \( \frac{1}{11} \) and \( \frac{11}{72} \), which is 792.Convert \( \frac{1}{11} \) and \( \frac{11}{72} \) to have denominator 792:\[\frac{1}{11} = \frac{72}{792},\quad \frac{11}{72} = \frac{121}{792}\]Add them:\[\frac{72}{792} + \frac{121}{792} = \frac{193}{792}\]Thus, the probability that the noted number is either 7 or 8 is \(\frac{193}{792}\).

Step 6: Compare with Given Options

The calculated probability is \( \frac{193}{792} \), which matches option (A) \( \frac{193}{792}\). Hence, the answer is (A).

Key Concepts

Coin Toss

A coin toss is one of the simplest events in probability. An unbiased coin - one with equal likelihood of showing heads or tails - presents a straightforward scenario. When you flip such a coin, the sample space consists of just two possible outcomes: heads (H) and tails (T). Each outcome has an equal probability of \( \frac{1}{2} \).

In practice, a coin toss is often used to decide between two alternatives. However, in probability exercises like this one, it forms part of a larger system of random events. Understanding this simple concept is foundational to exploring more complex probability situations.

Sample Space

The idea of a sample space is crucial in probability. It refers to the set of all possible outcomes of a probabilistic experiment. For the exercise, there are two key sample spaces.

• The sample space for tossing a coin: \( \{ H, T \} \).
• The sample space for rolling two dice consists of 36 outcomes as each die can land in one of six ways. Thus, \(6 \times 6 = 36 \).
  e.g., (1,1), (1,2), ..., (6,6).

When you extend sample spaces to complex scenarios, such as outcomes of multiple events (like the head leading to dice throws), understanding interactions between these basic sample spaces becomes vital. Always think of sample space as the collection of what could possibly occur, forming the basis on which probabilities are assigned to events.

Dice Rolling

Rolling two dice adds another layer to probability exercises. Each die has six faces, numbered from 1 to 6. Combining the outcomes creates a grid of possibilities.

• A pair could sum to 7 through several combinations: (1,6), (2,5), (3,4), (4,3), (5,2), (6,1).
• Similarly, combinations for a sum of 8 are: (2,6), (3,5), (4,4), (5,3), (6,2).

Determining these combinations is essential to calculating specific probabilities. The more favorable outcomes there are, the higher the probability of that sum. In the given scenario, to find the probability of summing to either 7 or 8, count the number of successful outcomes and divide by the total number of combinations (36).
Each step in the process assists in revealing how seemingly random events adhere to predictable patterns and frequencies.

Card Picking

Card picking introduces replaced probabilistic outcomes to our exercise. From an ordered set of cards numbered 2 to 12, you select one without knowing its face. Since it involves choosing one of the 11 options, each number in the range has an equal probability of \( \frac{1}{11} \).

This exercise requires finding the chance of drawing cards with specific numbers - either a 7 or 8. Since both numbers appear exactly once, the probability of picking either is straightforward: \( \frac{2}{11} \).

In probability, knowing how to calculate such outcomes involves combining this approach with others, maximizing your odds understanding by blending separate probabilities effectively. How random elements interact contributes significantly to your grasp of probability as a whole.