Consider three biased coins. Let the probability of getting head be $\frac{1}{3}$ and the probabilit...

Given the probabilities for each coin:$ P(\text{Head}) = \frac{1}{3} $$ P(\text{Tail}) = \frac{2}{3} $Let $ P(F) $ be the probability that the first coin shows a tail:$ P(F) = \frac{2}{3} $To find the conditional probability $ P(E \mid F) $, where $ E $ is the event "at least two heads show up", we use the formula:$ P(E \mid F) = \frac{P(E \cap F)}{P(F)} $Here, $ P(E \cap F) $ is the probability that the first coin is a tail and at least two heads appear in the three tosses. With the first coin being a tail, we focus on the last two coins having at least two heads, which happens only if both are heads:$ P(\text{HH}) = \frac{1}{3} \times \frac{1}{3} = \frac{1}{9} $Thus, the probability of $ E \cap F $ is:$ P(E \cap F) = P(F) \times P(\text{HH}) = \frac{2}{3} \times \frac{1}{9} = \frac{2}{27} $Plugging this into the formula for conditional probability gives:$ P(E \mid F) = \frac{P(E \cap F)}{P(F)} = \frac{\frac{2}{27}}{\frac{2}{3}} = \frac{2}{27} \times \frac{3}{2} = \frac{1}{9} $Therefore, the conditional probability $ P(E \mid F) $ is:$ \frac{1}{9} $

A ship sets off on a voyage with three engines, labelled $A, B$, and $C$, which work independently. ...

Probability that engine $A$ breaks down, $P(A_{\text{break}}) = \frac{1}{4}$Probability that engine $B$ breaks down, $P(B_{\text{break}}) = \frac{1}{4}$Probability that engine $C$ breaks down, $P(C_{\text{break}}) = \frac{1}{2}$From these, calculate the probabilities that the engines work:$P(A_{\text{work}}) = 1 - P(A_{\text{break}}) = 1 - \frac{1}{4} = \frac{3}{4}$$P(B_{\text{work}}) = 1 - P(B_{\text{break}}) = 1 - \frac{1}{4} = \frac{3}{4}$$P(C_{\text{work}}) = 1 - P(C_{\text{break}}) = 1 - \frac{1}{2} = \frac{1}{2}$The ship cannot complete the voyage if:All three engines fail.Exactly two engines fail.Case 1: All three engines fail$ P(A_{\text{break}} \cap B_{\text{break}} \cap C_{\text{break}}) = P(A_{\text{break}}) \times P(B_{\text{break}}) \times P(C_{\text{break}}) = \frac{1}{4} \times \frac{1}{4} \times \frac{1}{2} = \frac{1}{32} $Case 2: Exactly two engines failIf $A$ and $B$ fail, and $C$ works:$ = \frac{1}{4} \times \frac{1}{4} \times \frac{1}{2} = \frac{1}{32} $If $A$ and $C$ fail, and $B$ works:$ = \frac{1}{4} \times \frac{3}{4} \times \frac{1}{2} = \frac{3}{32} $If $B$ and $C$ fail, and $A$ works:$ = \frac{3}{4} \times \frac{1}{4} \times \frac{1}{2} = \frac{3}{32} $Total Probability of Failure:$ = \frac{1}{32} + \frac{1}{32} + \frac{3}{32} + \frac{3}{32} = \frac{8}{32} = \frac{1}{4} $Probability that the Ship Can Complete the Voyage:$ P(\text{complete}) = 1 - P(\text{fail}) = 1 - \frac{1}{4} = \frac{3}{4} $

Probability - IAT-IISER Maths PYQs

Q: A bag contains 8 balls, of which 5 are green and 3 are red. The probability that two balls chosen at...

To determine the probability that two balls chosen at random from a bag are of the same color, we solve it as follows:Total number of ways to choose 2 balls out of 8:$ = { }^8 C_2 = \frac{8!}{2!(8-2)!} = \frac{8 \times 7}{2 \times 1} = 28 $Number of ways to choose 2 green balls out of 5:$ = { }^5 C_2 = \frac{5!}{2!(5-2)!} = \frac{5 \times 4}{2 \times 1} = 10 $Number of ways to choose 2 red balls out of 3:$ = { }^3 C_2 = \frac{3!}{2!(3-2)!} = \frac{3 \times 2}{2 \times 1} = 3 $Probability that both balls are of the same color:$ \text{Probability} = \frac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}} = \frac{10 + 3}{28} = \frac{13}{28} $Thus, the probability that two balls chosen at random are of the same color is $\frac{13}{28}$.

Q: A randomly chosen card from a deck of 52 cards is given to be a black card (Spade or Club). What is ...

$$ n(A \cup B)=n(A)+n(B)-n(A \cap B) $$Number of cards $=$ (Number of black face cards) + (Number of Spades)(Number of Spade face cards)Number of cards $=6+13-3=16$$$ \text { Probability }=\frac{\text { Number of favorable outcomes }}{\text { Total number of black cards }}=\frac{16}{26} $$Probability $=\frac{8}{13}$

Q: Five fair coins are tossed independently. What is the probability that at least two heads appear?...

We toss5 fair coinsindependently.We are asked for the probability thatat least two headsappear.Step 1. Total possible outcomesEach coin has 2 outcomes → total outcomes = $ 2^5 = 32 $.Step 2. Find the complementIt’s often easier to use the complement:$ P(\text{at least 2 heads}) = 1 - P(\text{fewer than 2 heads}) $“Fewer than 2 heads” means0 headsor1 head.Step 3. Compute $ P(0 \text{ heads}) $ and $ P(1 \text{ head}) $$ P(0 \text{ heads}) = \frac{\binom{5}{0}}{2^5} = \frac{1}{32} $$ P(1 \text{ head}) = \frac{\binom{5}{1}}{2^5} = \frac{5}{32} $So,$ P(\text{fewer than 2 heads}) = \frac{1 + 5}{32} = \frac{6}{32} = \frac{3}{16} $Step 4. Compute the desired probability$ P(\text{at least 2 heads}) = 1 - \frac{3}{16} = \frac{13}{16} $✅Final Answer:$ \boxed{\frac{13}{16}} $Correct Option:A

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PYQ 2020

medium

maths ID: iat-iise

A bag contains 8 balls, of which 5 are green and 3 are red. The probability that two balls chosen at random are of the same colour is :

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02

PYQ 2022

easy

maths ID: iat-iise

A randomly chosen card from a deck of 52 cards is given to be a black card (Spade or Club). What is the probability that it is either a face card (King, Queen or Jack) or a Spade?

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9/13

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PYQ 2023

easy

maths ID: iat-iise

Consider three biased coins. Let the probability of getting head be and the probability of getting tail be in each of the coins. Consider the experiment of tossing the threee coins one by one, and the following events: :"At least two heads show up"F:"First coin shows tail".What is the conditional probability of given that has already occurred?

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PYQ 2024

medium

maths ID: iat-iise

A ship sets off on a voyage with three engines, labelled, and, which work independently. The ship can complete the voyage only if at least two of these engines keep working. The probability that engine breaks down is, that engine breaks down is, and that engine breaks down is . What is the probability that the ship can complete the voyage?

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3/4

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05

PYQ 2025

easy

maths ID: iat-iise

Five fair coins are tossed independently. What is the probability that at least two heads appear?

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4

Probability

High-Yield Trend

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Probability

High-Yield Trend

Questions 5 MCQs

Official Solution

Official Solution

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5 MCQs