Tree diagram allow us to see visually all possible outcomes of an random experiment. Each branch in a tree diagram represent a possible outcome.
(i) When we throw a die, then from the tree diagram the sample space can be written as
S = {1, 2, 3, 4, 5, 6 }
When we toss two coins, then fr om the tree diagram the sample space can be written as
S = {HH, HT, TH, TT}
Example 1 :
Write the sample space for tossing three coins using tree diagram.
Solution :
The required sample space
= {HHH. HHT, HTH, HTT, THH, THT, TTH, TTT}
Total number of outcomes = 8.
Example 2 :
Write the sample space for selecting two balls from a bag containing 6 balls numbered 1 to 6 (using tree diagram).
Solution :
S = {(1, 1) (1, 2) (1, 3) (1, 4) (1, 5) (1, 6) (2, 1) (2, 2) (2, 3) (2, 4) (2, 5) (2, 6) (3, 1) (3, 2) (3, 3) (3, 4) (3, 5) (3, 6) (4, 1) (4, 2) (4, 3) (4, 4) (4, 5) (4, 6) (5, 1) (5, 2) (5, 3) (5, 4) (5, 5) (5, 6) (6, 1) (6, 2) (6, 3) (6, 4) (6, 5) (6, 6)}
Total number of outcomes = 36
Example 3 :
If A is an event of a random experiment such that P(A) : P (A bar) = 17 : 15 and n(S) = 640 then find (i) P(A bar ) (ii) n(A).
Solution :
P(A) : P (A bar) = 17 : 15
P(A) / P (A bar) = 17 / 15
P(A) / [1 - P (A)] = 17 / 15
15 P(A) = 17 [1 - P(A)]
15 P(A) = 17 - 17 P(A)
15 P(A) + 17 P(A) = 17
32 P(A) = 17
P(A) = 17/32
P(A) = n(A) / n(S)
In order to make the denominator of p(A), let us multiply by 20
P(A) = (17/32) ⋅ (20/20)
P(A) = 340/640
P(A bar) = 1 - P(A)
= 1 - (340/640)
= 300/640
P(A bar) = 15/32
n(A) = 340
Example 4 :
A coin is tossed thrice. What is the probability of getting two consecutive tails?
Solution :
The required sample space
= {HHH. HHT, HTH, HTT, THH, THT, TTH, TTT}
n(S) = 8
Let "A" be the event of getting two consecutive tails
A = { HTT, TTH, TTT }
n(A) = 3
P(A) = n(A) / n(S)
P(A) = 3/8
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