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If a coin is tossed twice what is the probability that same face turns up both times?
Answer: A
Total possible outcomes: HH, HT, TH, TT i.e., 4
⇒ Desired outcomes: HH, TT i.e., 2
Required probability = 2/4 = 1/2
Hence, option (a).
If a coin is tossed thrice what is the probability that at least one head shows up?
Answer: B
P (at least one heads) = 1 – P (all tails)
∴ P (all tails) = 1/2 × 1/2 × 1/2 = 1/8
P (at least one heads) = 1 – 1/8 = 7/8
Hence, option (b).
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 three heads.
Answer: C
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.
P (3 heads) = 21/175 = 0.12
Hence, option (c).
Suppose a fair coin is randomly tossed for 75 times and it is found that head turns up 45 times and tail 30 times. What is the probability of getting a head?
0.6
Answer: A
Total number of trials = 75.
Number of times head turns up = 45
Number of times tail turns up = 30
P (head) = 45/75 = 0.6
Hence, option (a).
A person tosses a coin. If head comes up he gets Rs. 20 but loses Rs. 10 if tail comes up. What is his expected value?
Answer: 5
If head comes up he gets Rs. 20 but loses Rs. 10 if tail comes up
P (head) = P (tail) = ½
Expected value = 1/2 × 20 - 1/2 × 10 = Rs. 5.
Hence, 5.
If a coin is tossed thrice what is the probability of getting exactly 2 heads?
Answer: C
Total possible outcomes = 2 × 2 × 2 = 8
Desired outcomes: 2 heads and 1 tail can be arranged in 3!/2! = 3 ways.
∴ P (exactly 2 heads) = 3/8
Hence, option (c).
If a coin is tossed four times what is the probability that same face does not show up in any two consecutive tosses?
Answer: A
Total possible outcomes = 2 × 2 × 2 × 2 = 16
Desired outcomes = HTHT or THTH i.e., 2.
∴ Required probability = 2/16 = 1/8
Hence, option (a).
An unbiased coin is tossed a fixed number of times. If the probability of getting 5 heads equals the probability of getting 6 heads, then what is the probability of getting 1 head?
Answer: C
Let the coin is tossed n times.
P (5 heads) = nC5 × (1/2)5 × (1/2)n-5
P (6 heads) = nC6 × (1/2)6 × (1/2)n-6
∴ nC5 × (1/2)5 × (1/2)n-5 = nC6 × (1/2)6 × (1/2)n-6
⇒
⇒ 6 = n - 5
⇒ n = 11
∴ P (1 head) = 11C1 × (1/2)1 × (1/2)10 = 11/2048.
Hence, option (c).
In a biased coin, head occurs twice as frequently as tail occurs. If the coin is tossed 3 times, what is the probability of getting two heads?
Answer: B
Given, P(H) = 2 × P(T)
Also, P(H) + P(T) = 1
⇒ 2P(T) + P(T) = 1
∴ P(T) = 1/3 and P(H) = 2/3
Now, P(2 heads out of 3 tosses) = 3C2 × (2/3)2 × (1/3) = 3 × 4/27 = 12/27
Hence, option (b).