Two fair coins are tossed. What is the probability that at least one is a head? (Enter the probability as a fr?
Possible outcomes: 2 heads....has at least one head 1 head, 1 tail...has at least one head 1 tail, 1 head...has at least one head 2 tails...does not have at least one head P(at least one head in two tosses) = 3/4
A coin is tossed twice.what is the probability that at least 1 head occurs?
3/4 coz no. of possible outcomes when a coin is tossed are...... heads heads heads tails tails heads tails tails no. of favourable outcomes=3(when atleast one head occur coz there can be more than one also) total outcomes=4 probability=3/4
Three coins are tossed. What is the probability of at least one head?
First, suppose we have a fair coin (i.e probabilities of getting head or tail are 0.5 for both). Let [math]A[/math] be the event, where we get at least one head. We know that [math]P(A)=1-P(A^c)[/math] , where [math]A^C[/math] is the opposite event (no heads). Probability of getting no heads is[math] (\frac{1}{2})^3=\frac{1}{8}. [/math]So [math]P(A)=1-\frac{1}{8}=\frac{7}{8}.[/math]The probability of getting at most one coin is equal to sum of probalities that we get no coins and at most one coin, which are [math]\frac{1}{8}[/math] and [math]3\cdot\frac{1}{8} [/math](since we can choose three throws at which we get a coin). So, our total probability of that event is [math]\frac{1+3}{8}=\frac{4}{8}=\frac{1}{2}.[/math]
A pair of coins is tossed once, What is the probability of at least one head showing?
Always easier to work this type of problem "backwards".P (at least one head) = 1 - P(no heads)=1 - P(all tails)=1 - (1/2 X 1/2)=1 - 1/4=0.75
Two fair coins are tossed. What is the probability of 2 tails, given at least one is a tail?
For any two events A and B, with P(B) ≠ 0, then the conditional probability of A given B is: P(A | B) = P( A ∩ B) / P(B) Let A be the event of two tails Let B be the event at least one tail Possible outcomes are: HH, HT, TH, TT P(A ∩ B ) = P( HH ) = 1/4 P(B) = P( HT, TH, TT) = 3/4 P( A | B ) = (1/4) / (3/4) = 1/3
A coin with P (Head) =0.6 is tossed twice. What is the probability that at least one head occurs?
P(Heads) = 0.6 P(Tails) = 0.4 P(at least 1 head) = 1 − P(2 tails) = 1 − (0.4)² = 0.84
Two fair coins are tossed once. What is the probability of getting at least one heads?
All outcome for 2 coins:h,hh,tt,ht,t3 out of 4 possibilities have at least one head, so the chance is 3/4, or 75%
Two coins are tossed. What is the probability of coming up with two heads if it is known that at least one head comes up?
Comes under conditional probabailitySo given atleast one head occurs~{H,T} {T,H} {H,H}P(E) = n(E)/n(S) = 1/3