# CRE 5 - Coin Toss | Modern Math - Probability

**CRE 5 - Coin Toss | Modern Math - Probability**

If a coin is tossed twice what is the probability that same face turns up both times?

- A.
1/2

- B.
1/4

- C.
1

- D.
None of these

Answer: Option A

**Explanation** :

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).

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**CRE 5 - Coin Toss | Modern Math - Probability**

If a coin is tossed thrice what is the probability that at least one head shows up?

- A.
1/8

- B.
7/8

- C.
3/4

- D.
None of these

Answer: Option B

**Explanation** :

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).

Workspace:

**CRE 5 - Coin Toss | Modern Math - Probability**

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.

- A.
0.22

- B.
0.35

- C.
0.12

- D.
None of these

Answer: Option C

**Explanation** :

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).

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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?

**CRE 5 - Coin Toss | Modern Math - Probability**

0.6

- A.
0.5

- B.
0.4

- C.
None of these

Answer: Option A

**Explanation** :

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).

Workspace:

**CRE 5 - Coin Toss | Modern Math - Probability**

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

**Explanation** :

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.

Workspace:

**CRE 5 - Coin Toss | Modern Math - Probability**

If a coin is tossed thrice what is the probability of getting exactly 2 heads?

- A.
1/8

- B.
7/8

- C.
3/8

- D.
None of these

Answer: Option C

**Explanation** :

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).

Workspace:

**CRE 5 - Coin Toss | Modern Math - Probability**

If a coin is tossed four times what is the probability that same face does not show up in any two consecutive tosses?

- A.
1/8

- B.
7/8

- C.
3/8

- D.
None of these

Answer: Option A

**Explanation** :

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).

Workspace:

**CRE 5 - Coin Toss | Modern Math - Probability**

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?

- A.
5/1024

- B.
13/2048

- C.
11/2048

- D.
None of these

Answer: Option C

**Explanation** :

Let the coin is tossed n times.

P (5 heads) = ^{n}C_{5} × (1/2)^{5} × (1/2)^{n-5}

P (6 heads) = ^{n}C_{6} × (1/2)^{6} × (1/2)^{n-6}

∴ ^{n}C_{5} × (1/2)^{5} × (1/2)^{n-5} = ^{n}C_{6} × (1/2)^{6} × (1/2)^{n-6}

⇒ $\frac{n!}{5!\left(n-5\right)!}=\frac{n!}{6!\left(n-6\right)!}$

⇒ 6 = n - 5

⇒ n = 11

∴ P (1 head) = ^{11}C_{1} × (1/2)^{1} × (1/2)^{10} = 11/2048.

Hence, option (c).

Workspace:

**CRE 5 - Coin Toss | Modern Math - Probability**

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?

- A.
3/8

- B.
4/27

- C.
12/27

- D.
None of these

Answer: Option B

**Explanation** :

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) = ^{3}C_{2} × (2/3)^{2} × (1/3) = 3 × 4/27 = 12/27

Hence, option (b).

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