Two coins are tossed 100 times, two tails came up 30 times, one tail came up 45 times and no tail turned up 25 times. If the coins are tossed again then the probability of getting 2 heads is
Answers
Answer:
Total number of possible outcomes = 2
(i) If the favourable outcome is head (H).
Number of favourable outcomes = 1.
Therefore, P(getting a head)
Number of favorable outcomes
= P(H) = total number of possible outcomes
= 1/2.
(ii) If the favourable outcome is tail (T).
Number of favourable outcomes = 1.
Therefore, P(getting a tail)
Number of favorable outcomes
= P(T) = total number of possible outcomes
= 1/2.
Word Problems on Coin Toss Probability:
1. A coin is tossed twice at random. What is the probability of getting
(i) at least one head
(ii) the same face?
Solution:
The possible outcomes are HH, HT, TH, TT.
So, total number of outcomes = 4.
(i) Number of favourable outcomes for event E
= Number of outcomes having at least one head
= 3 (as HH, HT, TH are having at least one head).
So, by definition, P(F) = 34.
(ii) Number of favourable outcomes for event E
= Number of outcomes having the same face
= 2 (as HH, TT are have the same face).
So, by definition, P(F) = 24 = 12.
2. If three fair coins are tossed randomly 175 times and it is found that three heads appeared 21 times, two heads appeared 56 times, one head appeared 63 times and zero head appeared 35 times.
What is the probability of getting
(i) three heads, (ii) two heads, (iii) one head, (iv) 0 head.
Solution:
Total number of trials = 175.
Number of times three heads appeared = 21.
Number of times two heads appeared = 56.
Number of times one head appeared = 63.
Number of times zero head appeared = 35.
Let E1, E2, E3 and E4 be the events of getting three heads, two heads, one head and zero head respectively.
(i) P(getting three heads)
Number of times three heads appeared
= P(E1) = total number of trials
= 21/175
= 0.12
(ii) P(getting two heads)
Number of times two heads appeared
= P(E2) = total number of trials
= 56/175
= 0.32
(iii) P(getting one head)
Number of times one head appeared
= P(E3) = total number of trials
= 63/175
= 0.36
(iv) P(getting zero head)
Number of times zero head appeared
= P(E4) = total number of trials
= 35/175
= 0.20
Note: Remember when 3 coins are tossed randomly, the only possible outcomes
are E2, E3, E4 and
P(E1) + P(E2) + P(E3) + P(E4)
= (0.12 + 0.32 + 0.36 + 0.20)
= 1