Ten coins are tossed. The probability of getting atleast 8 heads is: (a) (7)/(16) (b) (7)/(64) (c) (7)/(128) (

Question 72 of 89

Q. Ten coins are tossed. The probability of getting atleast 8 heads is:

(a)

\dfrac{7}{16}

(b)

\dfrac{7}{64}

(c)

\dfrac{7}{128}

(d)

\dfrac{7}{32}

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Concept understanding — Classical Probability

Classical Probability: The "Fair Game" Definition

Imagine you roll a fair six-sided die. Before it lands, you know there are exactly six possible outcomes — 1, 2, 3, 4, 5, or 6 — and you have no reason to believe any one face is more likely than another. That gut feeling of "all outcomes are equally likely" is the entire foundation of classical probability.

The Intuition

Classical probability was born from games of chance — dice, coins, cards. In these settings, the physical symmetry of the objects (a balanced die, a fair coin) guarantees that no outcome is favoured. So the probability of an event is simply:

Number of ways the event can happen, divided by the total number of possible outcomes.

If you want the chance of rolling an even number on a die, count the evens: 2, 4, 6 — that's 3 ways. Total outcomes: 6. So probability = $3/6 = 1/2$ .

This is the "counting" approach. It works beautifully when the underlying experiment is symmetric and finite.

The Precise Statement

P(E) = \frac{\text{Number of outcomes favourable to event } E}{\text{Total number of equally likely outcomes}}

This is called the classical definition (or a priori definition) of probability. It was formalised by Pierre-Simon Laplace in the 18th century.

Three conditions must hold for this definition to apply:

Finite sample space — there are only a fixed, countable number of possible outcomes.
Equally likely outcomes — each outcome has the same chance of occurring (the "fairness" condition).
Mutually exclusive outcomes — no two outcomes can happen at the same time.

Watch out

The biggest mistake students make is applying classical probability to situations where outcomes are not equally likely. For example: "I can either pass or fail the exam — two outcomes, so probability of passing is $1/2$ ." That's nonsense, because passing and failing are not equally likely. The die works only because the die is fair.

A Simple Example

Problem: A bag contains 3 red marbles and 2 blue marbles. You pick one marble at random. What is the probability it is red?

Step 1 — Identify the sample space: There are 5 marbles total. If the marbles are physically identical except for colour, and you pick without looking, each marble is equally likely to be chosen. So total outcomes = 5.

Step 2 — Count favourable outcomes: 3 marbles are red. So favourable outcomes = 3.

Step 3 — Apply the formula:

P(\text{red}) = \frac{3}{5}

That's it. No deeper theory needed for this case.

When Classical Probability Fails

Classical probability cannot handle: …

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With 10 fair coins, $P(\text{at least 8 heads})=\dfrac{{}^{10}C_8+{}^{10}C_9+{}^{10}C_{10}}{2^{10}}=\dfrac{7}{128}$ .

Each coin toss is a Bernoulli trial with $p=\dfrac12$ probability of heads, $n=10$ tosses. The number of heads $X$ follows a Binomial( $10,\tfrac12$ ) distribution:

$P(X=k) = {}^{10}C_k\left(\frac12\right)^{10}$

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Q.Ten coins are tossed. The probability of getting atleast 8 heads is:

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