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Q4:
Suppose that 90% of people are right-handed. What is the probability that at most 6 of a random sample of 10 people are right-handed?

Solution :

Step 1: Identifying the Distribution Model

The problem involves a fixed number of independent Bernoulli trials, where each trial (selecting a person) has two possible outcomes: right-handed (success) or not right-handed (failure). Since the probability of success is constant, we apply the Binomial Distribution formula:

$P(X = k) = \binom{n}{k} p^k q^{n-k}$

Where:

  • $n = 10$ (Total number of people in the sample)
  • $p = 0.9$ (Probability that a person is right-handed)
  • $q = 1 - p = 0.1$ (Probability that a person is not right-handed)
  • $X$ is the random variable representing the number of right-handed people.

Step 2: Defining the Target Probability

We are required to find the probability that at most 6 people are right-handed. This is expressed as:

$P(X \leq 6) = \sum_{k=0}^{6} \binom{10}{k} (0.9)^k (0.1)^{10-k}$

Due to the law of total probability, it is computationally more efficient to calculate the complement and subtract from 1:

$P(X \leq 6) = 1 - [P(X=7) + P(X=8) + P(X=9) + P(X=10)]$

Step 3: Calculating Individual Probabilities

We compute each term $P(X=k) = \binom{10}{k} (0.9)^k (0.1)^{10-k}$:

$k$ Calculation: $\binom{10}{k} (0.9)^k (0.1)^{10-k}$ Value
$7$ $\frac{10!}{7!3!} (0.9)^7 (0.1)^3 = 120 \times 0.4782969 \times 0.001$ $0.0573956$
$8$ $\frac{10!}{8!2!} (0.9)^8 (0.1)^2 = 45 \times 0.4304672 \times 0.01$ $0.1937102$
$9$ $\frac{10!}{9!1!} (0.9)^9 (0.1)^1 = 10 \times 0.3874205 \times 0.1$ $0.3874205$
$10$ $\frac{10!}{10!0!} (0.9)^{10} (0.1)^0 = 1 \times 0.3486784 \times 1$ $0.3486784$

Step 4: Final Summation and Subtraction

Summing the probabilities for $k > 6$:

$\sum_{k=7}^{10} P(X=k) = 0.0573956 + 0.1937102 + 0.3874205 + 0.3486784 = 0.9872047$

Subtracting from total probability [Per the Complement Rule]:

$P(X \leq 6) = 1 - 0.9872047 = 0.0127953$

Visual Representation of Distribution

0 10 Right-handed sample count (k)

Final Solution: The probability that at most 6 people are right-handed is approximately 0.0128.


More Questions from Class 12 Mathematics Probability Miscellaneous Exercise on Chapter 13


CBSE Solutions for Class 12 Mathematics Probability


Chapters in CBSE - Class 12 Mathematics


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