CameraIcon
CameraIcon
SearchIcon
MyQuestionIcon
MyQuestionIcon
1
You visited us 1 times! Enjoying our articles? Unlock Full Access!
Question

Two coins are tossed simultaneously. Find the probability of getting
(i) exactly 1 head (ii) at most 1 head (iii) at least 1 head.

Open in App
Solution

If we say that the coin is fair. Then getting a head or a tail have equal probabilities. So, P(H)=P(T)=1/2

Now, P(at most 1 head)=P(getting 0 heads or 1 head) = P(0 heads) + P(1 head) = P(T)*P(T)+P(H)*P(T)+P(T)*P(H)= 3/4

getting at least 1 head:

Let E6 = event of getting at least 1 head. Then,
E6 = {HT, TH, HH} and, therefore, n(E6) = 3.
Therefore, P(getting at least 1 head) = P(E6) = n(E6)/n(S) = ¾.

exactly 1 head
E = event of getting 1 head = {HT , TH} = 2
p(getting 1 head) = 2/4 = 1/2


flag
Suggest Corrections
thumbs-up
7
Join BYJU'S Learning Program
similar_icon
Related Videos
thumbnail
lock
Coin Toss Experimental Probability
MATHEMATICS
Watch in App
Join BYJU'S Learning Program
CrossIcon