NCERT Solutions for Class 9th Maths Chapter 7 Think and Reflect — Tree diagrams
Book page 169 Updated on2026-09-08
Q1.
Can you calculate the probability of getting one head and one tail?
Answer
Yes. Read it off the tree of Fig. 7.6, whose four paths are equally likely.
S = {HH, HT, TH, TT}, n(S) = 4 E = 'one head and one tail' = {HT, TH}, n(E) = 2 P(E) = 2/4 = 1/2 = 0.5 or 50%
Four equally likely paths. HT and TH are two different paths — that is why one head and one tail is twice as likely as two heads.
Why it happens: HT and TH are two different paths through the tree — head first then tail, and tail first then head. They give the same description but they are separate outcomes, so the event contains two of the four leaves, not one. This is why P(one of each) = 1/2 while P(HH) = 1/4: getting one of each is twice as likely as getting two heads.
The common wrong answer is 1/3, from the list "two heads, one of each, two tails". That list is a legitimate sample space — nothing is missing and nothing repeats — but its three members are not equally likely (1/4, 1/2, 1/4), so dividing by 3 is not allowed. The tree protects you from this by keeping the two orders apart.
Check it yourself: multiply along a path instead of counting leaves. P(HT) = 1/2 × 1/2 = 1/4 and P(TH) = 1/2 × 1/2 = 1/4; the two paths make up the event, so add them: 1/4 + 1/4 = 1/2. Multiply along a path, add across paths — that rule will carry you through the rest of the chapter.