JEE Main 2019 — Mathematics Probability & Statistics

Probability of success i.e. getting a $5$5 or $6=P(s)=\frac{2}{6}=\frac{1}{3}$6=P(s)=62​=31​.

So, probability of failure i.e. getting any number other than $5$5 or $6=P(f)=1-P(s)=1-\frac{1}{3}=\frac{2}{3}$6=P(f)=1−P(s)=1−31​=32​.

Let $X$X be the amount that a man wins in a maximum of three throws.

So, $X$X can take values :

Case I : First two throws are $5$5 or $6$6, game stops

$$\Rightarrow X=100\Rightarrow P(X=100)=\frac{1}{3}$$⇒X=100⇒P(X=100)=31​

Case II : First throw is other than $5$5 or $6$6 and second and third throws are $5$5 or $6$6, game stops

$$\Rightarrow X=50\Rightarrow P(X=50)=\frac{2}{3}\times \frac{1}{3}=\frac{2}{9}$$⇒X=50⇒P(X=50)=32​×31​=92​

Case III : First two throws are other than $5$5 or $6$6 and last throw is $5$5 or $6$6

$$\Rightarrow X=0\Rightarrow P(X=0)=\frac{2}{3}\times \frac{2}{3}\times \frac{1}{3}=\frac{4}{27}$$⇒X=0⇒P(X=0)=32​×32​×31​=274​

Case IV : All three throws are other than $5$5 or $6$6

$$\Rightarrow X=-150\Rightarrow P(X=-150)=\frac{2}{3}\times \frac{2}{3}\times \frac{2}{3}=\frac{8}{27}$$⇒X=−150⇒P(X=−150)=32​×32​×32​=278​

Therefore, expected gain or loss

$$=E(X)=100\times \frac{1}{3}+50\times \frac{2}{9}+0\times \frac{4}{27}-150\times \frac{8}{27}=0$$=E(X)=100×31​+50×92​+0×274​−150×278​=0