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Exercise 9.3 · Q1

Q.In a group of 100 sports car buyers, 40 bought alarm systems, 30 purchased bucket seats, and 20 purchased an alarm system and bucket seats. If a car buyer chosen at random, bought an alarm system, what is the probability they also bought bucket seats?

Lakshadweep CbseNCERTSubjective· 2mImportance★★★★★est
48% · 10/21 Questions
✓ Free question

Applying the conditional-probability formula directly to the given counts gives P(bucket seats∣alarm)=12P(\text{bucket seats}\mid\text{alarm})=\dfrac12.

P(B∣A)=P(A∩B)P(A)=n(A∩B)n(A)P(B\mid A)=\frac{P(A\cap B)}{P(A)}=\frac{n(A\cap B)}{n(A)}

where A=A= "bought alarm system", B=B= "bought bucket seats", using counts since all buyers are equally likely to be chosen.

  1. Given data (out of 100100 buyers).

n(A)=40,n(B)=30,n(A∩B)=20n(A)=40,\qquad n(B)=30,\qquad n(A\cap B)=20

  1. Convert to probabilities.

P(A)=40100=0.4,P(A∩B)=20100=0.2P(A)=\frac{40}{100}=0.4,\qquad P(A\cap B)=\frac{20}{100}=0.2

  1. Apply the conditional probability formula.

P(B∣A)=P(A∩B)P(A)=0.20.4=12P(B\mid A)=\frac{P(A\cap B)}{P(A)}=\frac{0.2}{0.4}=\frac12

  1. Equivalently, working purely with counts among the 4040 alarm-buyers: 2020 of them also bought bucket seats, so P(B∣A)=2040=12P(B\mid A)=\dfrac{20}{40}=\dfrac12.

Self-check: Both the probability-ratio method and the direct-count method give the identical value 12\frac12, confirming the computation.

✓Final answer

P(bucket seats∣alarm system)=12=0.5P(\text{bucket seats}\mid\text{alarm system})=\dfrac{1}{2}=0.5 (i.e. 50%50\%).

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