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NCERT Exemplar · Q10

Q.One of the four persons John, Rita, Aslam or Gurpreet will be promoted next month. Consequently the sample space consists of four elementary outcomes S={John promoted, Rita promoted, Aslam promoted, Gurpreet promoted}S = \{\text{John promoted, Rita promoted, Aslam promoted, Gurpreet promoted}\}. You are told that the chances of John's promotion is same as that of Gurpreet, Rita's chances of promotion are twice as likely as John's. Aslam's chances are four times that of John.

(a) Determine P(John promoted)P(\text{John promoted}), P(Rita promoted)P(\text{Rita promoted}), P(Aslam promoted)P(\text{Aslam promoted}), P(Gurpreet promoted)P(\text{Gurpreet promoted}).
(b) If A={John promoted or Gurpreet promoted}A = \{\text{John promoted or Gurpreet promoted}\}, find P(A)P(A).
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The problem uses the probability axioms (non‑negativity, sum = 1) to assign probabilities from given ratios.

Letting P(John)=xP(\text{John}) = x, we get P(Gurpreet)=xP(\text{Gurpreet}) = x, P(Rita)=2xP(\text{Rita}) = 2x, P(Aslam)=4xP(\text{Aslam}) = 4x.

Summing: x+x+2x+4x=8x=1⇒x=18x + x + 2x + 4x = 8x = 1 \Rightarrow x = \frac{1}{8}.

Then P(A)=P(John)+P(Gurpreet)=18+18=14P(A) = P(\text{John}) + P(\text{Gurpreet}) = \frac{1}{8} + \frac{1}{8} = \frac{1}{4}.


Why the probability axioms are the natural tool

The sample space has exactly four outcomes — mutually exclusive and exhaustive. The only rule we have is that the four probabilities must be non‑negative and add up to 1. The problem gives us relative chances, not absolute numbers. That’s a perfect cue to introduce a variable for the smallest unknown probability and express everything else in terms of it. The axioms then let us solve for that variable.

Watch out

A common mistake is to treat “twice as likely” as if it means P(Rita)=2×P(John)P(\text{Rita}) = 2 \times P(\text{John}) and also P(John)+P(Rita)=1P(\text{John}) + P(\text{Rita}) = 1 — that would ignore the other two persons. Always account for the entire sample space.


Step‑by‑step solution

  1. Choose a variable for the smallest probability.

    John and Gurpreet have the same chance. Rita’s chance is twice John’s; Aslam’s is four times John’s. So John’s probability is the natural base.

    Let P(John)=xP(\text{John}) = x. Then:

    • P(Gurpreet)=xP(\text{Gurpreet}) = x
    • P(Rita)=2xP(\text{Rita}) = 2x
    • P(Aslam)=4xP(\text{Aslam}) = 4x
  2. Apply the axiom of total probability.

    Since these four events are the only possibilities and they are mutually exclusive:

P(John)+P(Rita)+P(Aslam)+P(Gurpreet)=1P(\text{John}) + P(\text{Rita}) + P(\text{Aslam}) + P(\text{Gurpreet}) = 1

Substituting:

x+2x+4x+x=1⇒8x=1x + 2x + 4x + x = 1 \quad\Rightarrow\quad 8x = 1

Hence:

x=18x = \frac{1}{8}

  1. Write down each probability.

    • P(John)=18P(\text{John}) = \frac{1}{8}
    • P(Gurpreet)=18P(\text{Gurpreet}) = \frac{1}{8} …

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