A defective die is thrown ten times independently. The probability that an even number will appear 5 times is twice the probability that an even number will appear 4 times. What is the probability that odd face appear in each of the 10 throws.
Answers
Through an oversight I have considered a coin instead of a die. My sincere thanks to Ramakrushna Mishra for pointing this out.
In 1010 independent rollings of a biased die, the probability that an even number will appear is twice the probability that an even number will appear four times. What is the probability that an even number will appear twice when the die is rolled 88 times.
Let the probability of an even number appearing in a single throw of the biased die be p.p.
⇒⇒ The probability that all the 1010 throws will result in an odd number is (1−p)10.(1−p)10.
⇒⇒ The probability that in 1010 throws, an even number will appear at least once is 1−(1−p)10.1−(1−p)10.
The number of ways in which an even number will appear four times in 1010 throws is
C(10,4)p4(1−p)6=10!4!6!p4(1−p)6=210p4(1−p)6.C(10,4)p4(1−p)6=10!4!6!p4(1−p)6=210p4(1−p)6.
The total number of outcomes in 1010 throws is 210.210.
The probability that in 1010 throws, an even number will appear four times is
210p4(1−p)6210.210p4(1−p)6210.
It is given that in 1010 independent rollings of a biased die, the probability that an even number will appear is twice the probability that an even number will appear four times.
⇒1−(1−p)10=2×210p4(1−p)6210.⇒1−(1−p)10=2×210p4(1−p)6210.
⇒256(1−p)10+105p4(1−p)6−256=0.⇒256(1−p)10+105p4(1−p)6−256=0.
The value of pp satisfying this equation is about 2.505×10−17,2.505×10−17, which is so low that for all practical purposes it can be considered as zero.