Probability
Probability
Allen Star Batch
Grade 12
Question:
A certain coin is tossed with probability of showing head being 'p'. Let 'q' denote the probability that when the coin is tossed four times the number of heads obtained is even. Then:
There is no value of $p$, if $q = \frac{1}{4}$
There is exactly one value of $p$ if $q = \frac{3}{4}$
There are exactly two values of $p$ if $q = \frac{3}{5}$
There are exactly four values of $p$ if $q = \frac{4}{5}$
Step-by-Step Solution
Key Concept: The probability of getting an even number of heads in 4 tosses is q = (2p-1)⁴/2 + 1/2, which can be rearranged as 2q - 1 = (2p-1)⁴. For a given value of q, solving (2p-1)⁴ = 2q-1 determines how many real solutions p ∈ [0,1] exist, since p = (u+1)/2 where u⁴ = 2q-1.
We have $q = P(0H \text{ or } 2H \text{ or } 4H) = p^4 + ^4C_2p^2(1-p)^2 + (1-p)^4 = 8p^4 - 16p^3 + 12p^2 - 4p + 1 = \frac{(2p-1)^4 + 1}{2}$. Substituting the given value of $p$ and checking yields the required result.
Correct Answer: 1,3