Maths · Statistics and probability · chapter 14 of 14

Class 11 Maths Chapter 14: Probability case study question with answers

One case-based question of the kind CBSE-pattern school exams ask: a passage and a table, then 3 parts worth 4 marks in all, with the answer and the marking scheme for each. Written from the NCERT chapter.

Case study · 4 marks

A coach at a cricket academy in Hyderabad studies a medium-pace bowler. From many overs of match records she assigns a probability to each outcome of a ball, as in the table. Every ball has exactly one of these outcomes, so the outcomes form the sample space, and the probabilities must satisfy the axioms of probability. One value in the table, k, was smudged in her notebook.

Outcome	dot ball	1 run	2 runs	4 runs	6 runs	wicket
Probability	0.4	0.25	0.1	k	0.05	0.08

Part 1 · 1 mark

The value of k is

(a) 0.2(b) 0.08(c) 0.12(d) 0.88

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Answer: (c) 0.12

The probabilities of all outcomes add to 1: k = 1 – 0.88 = 0.12.

Part 2 · 1 mark

The probability that a ball yields at least 2 runs is

(a) 0.27(b) 0.15(c) 0.52(d) 0.17

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Answer: (a) 0.27

P(2 runs) + P(4 runs) + P(6 runs) = 0.1 + 0.12 + 0.05 = 0.27.

Part 3 · 2 marks

Find the probability that a ball is neither a boundary (4 or 6 runs) nor a wicket. Are the events 'a boundary' and 'a wicket' mutually exclusive?

Answer and marking scheme

A boundary and a wicket cannot happen on the same ball here (each ball has exactly one outcome), so the events are mutually exclusive and P(boundary or wicket) = 0.12 + 0.05 + 0.08 = 0.25. The probability of neither is 1 – 0.25 = 0.75.

the events are mutually exclusive, P(boundary or wicket) = 0.25 (1)
P(neither) = 0.75 (1)

Find the probability that a ball yields an odd number of runs or a wicket, giving a reason for the rule you use.

Answer and marking scheme

An odd number of runs here means 1 run (no other odd score appears in the table). '1 run' and 'wicket' are different outcomes of the same ball, so they are mutually exclusive and the probabilities add: 0.25 + 0.08 = 0.33.

odd runs = 1 run; the two events are mutually exclusive (1)
P = 0.33 (1)

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