Question -
Answer -
Given head is 3 timesas likely to occur as tail.
Now, let theprobability of getting a tail in the biased coin be x.
⇒ P (T) = x
And P (H) = 3x
For a biased coin, P(T) + P (H) = 1
⇒ x + 3x = 1
⇒ 4x = 1
⇒ x = 1/4
Hence, P (T) =1/4 and P (H) = 3/4
As the coin is tossedtwice, so the sample space is {HH, HT, TH, TT}
Let X be a randomvariable representing the number of tails.
Clearly, X can takethe value of 0, 1 or 2.
P(X = 0) = P (no tail)= P (H) × P (H) = ¾ × ¾ = 9/16
P(X = 1) = P (onetail) = P (HT) × P (TH) = ¾. ¼ × ¼. ¾ = 3/8
P(X = 2) = P (twotail) = P (T) × P (T) = ¼ × ¼ = 1/16
Hence, the requiredprobability distribution is,
| X | 0 | 1 | 2 |
| P (x) | 9/16 | 3/8 | 1/16 |