Three coins are tossed. What is the probability of getting (i) neither...
Solution:
When three coins are tossed, there are 2^3 = 8 possible outcomes.
The possible outcomes are:
- HHH
- HHT
- HTH
- THH
- TTH
- THT
- HTT
- TTT
(i) Neither 3 Heads nor 3 Tails:
The outcomes that satisfy this condition are HHT, HTH, THH, and THT. So, the probability of getting neither 3 Heads nor 3 Tails is:
P = 4/8 = 1/2
Therefore, option 'A' is incorrect.
Explanation of option 'D':
The probability of getting 3 Heads is 1/8 and the probability of getting 3 Tails is also 1/8. So, the probability of getting either 3 Heads or 3 Tails is:
P(3 Heads or 3 Tails) = P(3 Heads) + P(3 Tails) = 1/8 + 1/8 = 1/4
The probability of not getting either 3 Heads or 3 Tails is:
P(neither 3 Heads nor 3 Tails) = 1 - P(3 Heads or 3 Tails)
= 1 - 1/4
= 3/4
Therefore, option 'D' is correct.
Explanation of option 'B':
The probability of getting 2 Heads and 1 Tail or 2 Tails and 1 Head is:
P(2 Heads and 1 Tail or 2 Tails and 1 Head) = P(HHT or HTH or THH or TTH or THT or HTT)
= 6/8
= 3/4
Therefore, option 'B' is incorrect.
Explanation of option 'C':
The probability of getting at least 2 Heads or at least 2 Tails is:
P(at least 2 Heads or at least 2 Tails) = P(2 Heads and 1 Tail or 3 Heads or 2 Tails and 1 Head or 3 Tails)
= P(HHT or HTH or THH or TTH or THT or HTT or HHH or TTT)
= 7/8
Therefore, option 'C' is incorrect.
Three coins are tossed. What is the probability of getting (i) neither...
(i) Probability of 3 heads = 1/8 Also, Probability of 3 tails =1/8.
Required probability = 1- (1/8 + 1/8) = 6/8 = 3/4.