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Exercise 9.2 · Q1

Q.Give a real-life example of the following: Independent Events and Dependent Events.

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✓ Free question

Independent events do not influence each other's probability; dependent events do — illustrated with one everyday example of each.

Two events A,BA,B are independent if

P(A∩B)=P(A)⋅P(B)⟺P(B∣A)=P(B)P(A\cap B)=P(A)\cdot P(B)\quad\Longleftrightarrow\quad P(B\mid A)=P(B)

They are dependent if

P(B∣A)≠P(B)so thatP(A∩B)=P(A)⋅P(B∣A)P(B\mid A)\neq P(B)\quad\text{so that}\quad P(A\cap B)=P(A)\cdot P(B\mid A)

where P(B∣A)P(B\mid A) is the conditional probability of BB given AA has occurred.

  1. Independent events — real-life example. Tossing a fair coin and rolling a fair die at the same time. Let A=A= "coin shows Head", B=B= "die shows 5". The die has no way of "knowing" what the coin did, so

P(A∩B)=P(A)⋅P(B)=12×16=112P(A\cap B)=P(A)\cdot P(B)=\frac12\times\frac16=\frac{1}{12}

which matches direct counting (11 favourable outcome out of 1212 total). Since P(B∣A)=P(B)=16P(B\mid A)=P(B)=\dfrac16, the events are independent.

  1. Dependent events — real-life example. Drawing two cards without replacement from a well-shuffled deck of 52 cards. Let A=A= "1st card is a King" (P(A)=452P(A)=\dfrac{4}{52}) and B=B= "2nd card is a King". If AA has occurred, only 3 Kings remain among 51 cards, so

P(B∣A)=351≠P(B)=452P(B\mid A)=\frac{3}{51}\neq P(B)=\frac{4}{52}

The 2nd draw's probability genuinely depends on the outcome of the 1st draw, so AA and BB are dependent.

Self-check: In example 1, P(A)P(B)=12⋅16=112P(A)P(B)=\frac12\cdot\frac16=\frac1{12} equals the true joint probability — confirms independence. In example 2, 351=117≠452=113\frac{3}{51}=\frac{1}{17}\neq\frac{4}{52}=\frac{1}{13} — confirms dependence.

✓Final answer

Independent: tossing a coin and rolling a die simultaneously (coin outcome does not affect die outcome).

Dependent: drawing two cards in succession without replacement (2nd draw's probability depends on the 1st draw).

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