Browse by Stream
QnA
Home
QnA
Go To Your Study Dashboard

Get Answers to all your Questions

header-bg qa
  • Home
  • Engineering
  • Two coins and <img alt="\mathrm{S}" src="https://entrancecorner.oncodecogs.com/gif.latex?%5Cmathrm%7B

Two coins \mathrm{R} and \mathrm{S} are tossed. The 4 joint events \mathrm{H}_{R} \mathrm{H}_{S}, \mathrm{~T}_{R} \mathrm{~T}_{S}, \mathrm{H}_{R} T_{S}, \mathrm{~T}_{R} \mathrm{H}_{S} have probabilities 0.28 ,0.18,0.30,0.24 respectively, where \mathrm{H} represents head and \mathrm{T} represents tail. Which one of the following is TRUE?

Option: 1

The coin tosses are independent


Option: 2

\mathrm{R} is fair,\mathrm{S} is not


Option: 3

\mathrm{S} is fair \mathrm{R} is not


Option: 4

The coin tosses are dependent


Answers (1)

best_answer

\mathrm{P}\left(\mathrm{H}_{R} \mathrm{H}_{S}\right)=0.28
\mathrm{P}\left(\mathrm{T}_{R} T_{S}\right)=0.18
\mathrm{P}\left(\mathrm{H}_{R} T_{S}\right)=0.30
\mathrm{P}\left(\mathrm{T}_{R} H_{S}\right)=0.24

Two events \mathrm{A} and \mathrm{B} are said to be independent if
\mathrm{P(A \cap B)=P(A) P(B)}
\mathrm{P\left(H_{R}\right)=P\left(H_{S}\right)=P\left(T_{R}\right)=P\left(T_{S}\right)=\frac{1}{2}}
So option (a) is wrong.

Now, if \mathrm{R}is fair and \mathrm{S} is not then
\mathrm{P\left(H_{R}\right) =P\left(T_{R}\right)=\frac{1}{2}}
\mathrm{P\left(H_{R} H_{S}\right) =\frac{1}{2} P\left(H_{S}\right)=0.28}
\mathrm{P\left(H_{S}\right) =0.56}
\mathrm{P\left(H_{R} T_{S}\right) =\frac{1}{2} P\left(T_{S}\right)=0.30 }
\mathrm{P\left(T_{S}\right) =0.60 }

\mathrm{P\left(H_{s}\right)+P\left(T_{S}\right)=1.61>1}
which is not possible.
\mathrm{\Rightarrow} Option (b) is wrong
Similarly option (c) is wrong.

Posted by

Divya Prakash Singh

View full answer

JEE Main high-scoring chapters and topics

Study 40% syllabus and score up to 100% marks in JEE

Similar Questions

Trending Articles/News

anam-khan

JEE Main 2026: JCECEB invites fresh applications for Jharkhand state merit list correction

Latest Question