JEE Main 2019 — Mathematics Probability & Statistics

The probability of getting a $4$4 in a throw of a die is $\frac{1}{6}$61​.

The probability of getting other than $4$4 i.e. $4$4 i.e. numbers 1,2,3,5&6 in a single throw of a die is $\frac{5}{6}$65​.

Total throws done $=5$=5.

For the event to be completed in ${5}^{th}$5th throw, ${4}^{th}$4th and ${5}^{th}$5th throw must be $4$4 . Also, ${3}^{rd}$3rd throw must be other than $4$4 i.e. $4$4. Hence, cases for the first two position can be :

Case I : ${1}^{st}$1st throw must be $4$4 and ${2}^{nd}$2nd throw must not be $4$4 i.e. $4$4

Thus, $P(4\overset{-}{4}\overset{-}{4}44)=\frac{1}{6}.\frac{5}{6}.\frac{5}{6}.\frac{1}{6}.\frac{1}{6}=\frac{25}{{6}^{5}}$P(44−4−44)=61​.65​.65​.61​.61​=6525​.

Case II : ${1}^{st}$1st throw must not be $4$4 i.e. $4$4 and ${2}^{nd}$2nd throw must not be $4$4 i.e. $4$4

Thus, $P(\overset{-}{4}\overset{-}{4}\overset{-}{4}44)=\frac{5}{6}.\frac{5}{6}.\frac{5}{6}.\frac{1}{6}.\frac{1}{6}=\frac{125}{{6}^{5}}$P(4−4−4−44)=65​.65​.65​.61​.61​=65125​.

Case III : ${1}^{st}$1st throw must not be $4$4 i.e. $4$4 and ${2}^{nd}$2nd throw must be $4$4

Thus, $P(\overset{-}{4}4\overset{-}{4}44)=\frac{5}{6}.\frac{1}{6}.\frac{5}{6}.\frac{1}{6}.\frac{1}{6}=\frac{25}{{6}^{5}}$P(4−44−44)=65​.61​.65​.61​.61​=6525​.

Therefore, probability that experiment will end in the fifth throw given that two fours occurred in succession will be

$$P(4\overset{-}{4}\overset{-}{4}44)+P(\overset{-}{4}4\overset{-}{4}44)+P(\overset{-}{4}\overset{-}{4}\overset{-}{4}44)$$P(44−4−44)+P(4−44−44)+P(4−4−4−44)

$$=\frac{25}{{6}^{5}}+\frac{25}{{6}^{5}}+\frac{125}{{6}^{5}}$$=6525​+6525​+65125​

$=\frac{175}{{6}^{5}}$=65175​.