# A fair coin is tossed four times and a person wins \[\operatorname{Re}1\] for each head and loses \[Rs.1.5\] for each tail that turns up. Let \[p\] be the probability of a person losing \[Rs3.50\] after \[4\] tosses. Find \[4p\]?

Last updated date: 21st Mar 2023

•

Total views: 306.9k

•

Views today: 3.85k

Answer

Verified

306.9k+ views

Hint: Find the number of times we see heads and tails on the coin in \[4\] tosses by forming a linear equation relating the number of tosses to the money earned after \[4\] tosses. Find the value of probability \[p\] by using the formula for calculating probability of independent events.

We have a fair coin which is tossed \[4\] times and a person wins \[\operatorname{Re}1\] for each head and loses \[Rs.1.5\] for each tail. We have to find the probability \[p\] of a person losing \[Rs.3.5\] after \[4\] tosses.

We will begin by calculating the number of heads and tails that occur in a series of \[4\] tosses such that a person loses \[Rs.3.5\] after \[4\] tosses.

Let’s assume that the number of times that a coin shows heads is \[x\]. As the total number of tosses is \[4\], the number of times a coin shows tails is \[4-x\]. Thus, the amount of money a person gains by getting \[x\] heads is \[Rs.x\times 1=Rs.x\]and the amount of money a person loses by getting \[4-x\] tails is \[Rs.\left( 4-x \right)\times 1.5=Rs.\left( 6-1.5x \right)\]. As the total loss after \[4\] tosses is \[Rs.3.5\], we have \[6-1.5x-x=3.5\]. Further simplifying the equation, we get \[2.5=2.5x\Rightarrow x=1\]. Thus, the number of heads is \[x=1\] and the number of tails is \[4-x=3\].

We will now find the probability of getting \[1\] heads and \[3\] tails in four tosses.

We know that probability of any event is the ratio of the number of favourable outcomes to the total number of possible outcomes.

Each time we toss a coin, the probability of getting heads or tails is \[=\dfrac{1}{2}\]. Also, the occurrence of heads or tails in each toss is independent of occurrence of heads or tails in other tosses.

We know that if two events \[A\] and \[B\] are independent, then we have \[P\left( A\cap B \right)=P\left( A \right)\times P\left( B \right)\].

As probability of getting one heads \[=\dfrac{1}{2}\] and probability of getting each tails \[=\dfrac{1}{2}\]and the events are independent, we have, the probability of getting \[1\] heads and \[3\] tails in four tosses \[=4\times \] probability of getting \[1\] heads \[\times \] probability of getting \[3\] tails \[=p\].

Thus, we have \[p=4\times \left( \dfrac{1}{2} \right)\times {{\left( \dfrac{1}{2} \right)}^{3}}=4\times {{\left( \dfrac{1}{2} \right)}^{4}}=\dfrac{1}{4}\].

Hence, the value of \[4p\] is \[4p=\dfrac{4}{4}=1\].

Note: Probability of any event describes how likely an event is to occur or how likely it is that a proposition is true. The value of probability of any event always lies in the range \[\left[ 0,1 \right]\] where having \[0\] probability indicates that the event is impossible to happen, while having probability equal to \[1\] indicates that the event will surely happen. We must remember that the sum of probability of occurrence of some event and probability of non-occurrence of the same event is always \[1\].

We have a fair coin which is tossed \[4\] times and a person wins \[\operatorname{Re}1\] for each head and loses \[Rs.1.5\] for each tail. We have to find the probability \[p\] of a person losing \[Rs.3.5\] after \[4\] tosses.

We will begin by calculating the number of heads and tails that occur in a series of \[4\] tosses such that a person loses \[Rs.3.5\] after \[4\] tosses.

Let’s assume that the number of times that a coin shows heads is \[x\]. As the total number of tosses is \[4\], the number of times a coin shows tails is \[4-x\]. Thus, the amount of money a person gains by getting \[x\] heads is \[Rs.x\times 1=Rs.x\]and the amount of money a person loses by getting \[4-x\] tails is \[Rs.\left( 4-x \right)\times 1.5=Rs.\left( 6-1.5x \right)\]. As the total loss after \[4\] tosses is \[Rs.3.5\], we have \[6-1.5x-x=3.5\]. Further simplifying the equation, we get \[2.5=2.5x\Rightarrow x=1\]. Thus, the number of heads is \[x=1\] and the number of tails is \[4-x=3\].

We will now find the probability of getting \[1\] heads and \[3\] tails in four tosses.

We know that probability of any event is the ratio of the number of favourable outcomes to the total number of possible outcomes.

Each time we toss a coin, the probability of getting heads or tails is \[=\dfrac{1}{2}\]. Also, the occurrence of heads or tails in each toss is independent of occurrence of heads or tails in other tosses.

We know that if two events \[A\] and \[B\] are independent, then we have \[P\left( A\cap B \right)=P\left( A \right)\times P\left( B \right)\].

As probability of getting one heads \[=\dfrac{1}{2}\] and probability of getting each tails \[=\dfrac{1}{2}\]and the events are independent, we have, the probability of getting \[1\] heads and \[3\] tails in four tosses \[=4\times \] probability of getting \[1\] heads \[\times \] probability of getting \[3\] tails \[=p\].

Thus, we have \[p=4\times \left( \dfrac{1}{2} \right)\times {{\left( \dfrac{1}{2} \right)}^{3}}=4\times {{\left( \dfrac{1}{2} \right)}^{4}}=\dfrac{1}{4}\].

Hence, the value of \[4p\] is \[4p=\dfrac{4}{4}=1\].

Note: Probability of any event describes how likely an event is to occur or how likely it is that a proposition is true. The value of probability of any event always lies in the range \[\left[ 0,1 \right]\] where having \[0\] probability indicates that the event is impossible to happen, while having probability equal to \[1\] indicates that the event will surely happen. We must remember that the sum of probability of occurrence of some event and probability of non-occurrence of the same event is always \[1\].

Recently Updated Pages

Calculate the entropy change involved in the conversion class 11 chemistry JEE_Main

The law formulated by Dr Nernst is A First law of thermodynamics class 11 chemistry JEE_Main

For the reaction at rm0rm0rmC and normal pressure A class 11 chemistry JEE_Main

An engine operating between rm15rm0rm0rmCand rm2rm5rm0rmC class 11 chemistry JEE_Main

For the reaction rm2Clg to rmCrmlrm2rmg the signs of class 11 chemistry JEE_Main

The enthalpy change for the transition of liquid water class 11 chemistry JEE_Main