John tossed a penny 4 times and Jessica tossed a nickel 4 times. What is the probability that at least one of them getting 4 Heads?

Question

John tossed a penny 4 times and Jessica tossed a nickel 4 times. What is the probability that at least one of them getting 4 Heads?

The total possible way, $n(S) = 2^4 \cdot 2^4$, as each of the toss can be landed as Head or Tail and we have 8 tosses in total (4 from John and 4 from Jessica).

However, I don't understand the question of finding the probability that at least one of them getting 4 Heads. Here's my initial approach.

P(John getting all tails) = P(Jessica getting all tails) = $\frac1{2^4}$

P(none of them getting 4 Heads) = P(John getting all tails) and P(Jessica getting all tails)

P(none of them getting 4 Heads) = $\frac1{2^4} \cdot \frac1{2^4} = \frac1{256}$

P(at least one of them getting 4 Heads) $= 1 -$ P(none of them getting 4 Heads)

P(at least one of them getting 4 Heads) $= 1 - \frac1{256} = \frac{255}{256}$

So, the probability is $\approx 0.996$, which makes no sense that they are getting at least 4 Heads almost every tosses.

Therefore, I kindly would like request to point out my misunderstanding in calculating probability.

Thank you in advance.

Answer 1

My mistake was in this line, where P(none of them getting 4 Heads) = P(John getting all tails) and P(Jessica getting all tails)

The above statement is wrong as none of them getting 4 Heads means that they can get some heads and tails as well.

So, my fixed solution is P(none of them getting 4 Heads) = P(John getting at least one Tail) and P(Jessica getting at least one Tail)

P(John getting at least one Tail) = 1 - P(John getting all Heads)

= 1 - $\frac1{2^4}$ = $\frac{15}{16}$

Similarly, P(Jessica getting at least one Tail) = $\frac{15}{16}$

Therefore, P(none of them getting 4 Heads) = $\frac{15}{16}$ $*$ $\frac{15}{16}$ = $\frac{225}{256}$

P(at least one of them getting 4 Heads) = 1 - P(none of them getting 4 Heads)

= 1 - $\frac{225}{256}$ = $\frac{31}{256}$

Answer 2

Your revised solution is correct.

Here is another approach. Let $E$ be the event that John obtains four heads; let $F$ be the event that Jessica obtains four heads. Then the probability that at least one of them gets four heads is $$\Pr(E \cup F) = \Pr(E) + \Pr(F) - \Pr(E \cap F)$$ since simply adding the probabilities that events $E$ and $F$ occur would count the event that they both occur twice. We only want to count it once, so we must subtract it from the total.

Assuming the coins are fair, the probability of obtaining a head on each toss is $1/2$. Since John tosses the penny four times, the probability that he obtains four heads is $$\Pr(E) = \left(\frac{1}{2}\right)^4$$ Since Jessica tosses the nickel four times, the probability that she obtains four heads is $$\Pr(F) = \left(\frac{1}{2}\right)^4$$ The probability that they both obtain four heads is $$\Pr(E \cap F) = \left(\frac{1}{2}\right)^4 \cdot \left(\frac{1}{2}\right)^4 = \left(\frac{1}{2}\right)^8$$ Hence, assuming the coins are fair, the probability that at least one of them obtains four heads is $$\Pr(E \cup F) = \left(\frac{1}{2}\right)^4 + \left(\frac{1}{2}\right)^4 - \left(\frac{1}{2}\right)^8 = \frac{1}{16} + \frac{1}{16} - \frac{1}{256} = \frac{16 + 16 - 1}{256} = \frac{31}{256}$$ as you found.